How to insert json data into my sql using php

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Posted by Valli Pandy On 12/08/2014
How to Insert JSON Data into MySQL using PHP
Hi, in this PHP TUTORIAL, we'll see How
to insert JSON Data into MySQL
using PHP. Check out its reverse
process of Converting Data from MySQL
to JSON Format in PHP here. Converting
json to mysql using php includes several
steps and you will learn things like how to
read json file, convert json to array and
insert that json array into mysql database
in this tutorial. For those who wonder
what is JSON, let me give a brief
introduction.
JSON file contains information stored in JSON format and has the extension of "*.json". JSON
stands for JavaScript Object Notation and is a light weight data exchange format. Being
less cluttered and more readable than XML, it has become an easy alternative format to store
and exchange data. All modern browsers supports JSON format.
Do you want to know how a JSON file looks like? Well here is the sample.
What is JSON File Format?
Example of a JSON File
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As you can see by yourself, the JSON format is very human readable and the above file
contains some employee details. I'm going to use this file as an example for this tutorial and
show you how to insert this JSON object into MySQL database in PHP step by step.
As the first and foremost step we have to connect PHP to the MySQL database in order to
insert JSON data into MySQL DB. For that we use mysql_connect() function to connect
PHP with MySQL.
<?php
$con=mysql_connect("username","password","")ordie('Couldnotconn
ect:'.mysql_error());
mysql_select_db("employee",$con);
?>
Here "employee" is the MySQL Database name we want to store the JSON object. Learn more
about using mysqli library for php and mysql database connection here.
Next we have to read the JSON file and store its contents to a PHP variable. But how to read
json file in php? Well! PHP supports the function file_get_contents() which will read
an entire file and returns it as a string. Let’s use it to read our JSON file.
<?php
//readthejsonfilecontents
$jsondata=file_get_contents('empdetails.json');
Step 1: Connect PHP to MySQL Database
Step 2: Read the JSON file in PHP
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?>
Here "empdetails.json" is the JSON file name we want to read.
The next step for us is to convert json to array. Which is likely we have to convert the JSON
string we got from the above step to PHP associative array. Again we use the PHP json
decode function which decodes JSON string into PHP array.
<?php
//convertjsonobjecttophpassociativearray
$data=json_decode($jsondata,true);
?>
The first parameter $jsondata contains the JSON file contents.
The second parameter true will convert the string into php associative array.
Next we have to parse the above JSON array element one by one and store them into PHP
variables.
<?php
//gettheemployeedetails
$id=$data['empid'];
$name=$data['personal']['name'];
$gender=$data['personal']['gender'];
$age=$data['personal']['age'];
$streetaddress=$data['personal']['address']['streetaddress'];
$city=$data['personal']['address']['city'];
$state=$data['personal']['address']['state'];
$postalcode=$data['personal']['address']['postalcode'];
$designation=$data['profile']['designation'];
$department=$data['profile']['department'];
?>
Using the above steps, we have extracted all the values from the JSON file. Finally let's insert
the extracted JSON object values into the MySQL table.
<?php
//insertintomysqltable
$sql="INSERTINTOtbl_emp(empid,empname,gender,age,streetaddres
s,city,state,postalcode,designation,department)
VALUES('$id','$name','$gender','$age','$streetaddress','$city',
'$state','$postalcode','$designation','$department')";
if(!mysql_query($sql,$con))
Step 3: Convert JSON String into PHP Array
Step 4: Extract the Array Values
Step 5: Insert JSON to MySQL Database with PHP Code

{
die('Error:'.mysql_error());
}
?>
We are done!!! Now we have successfully imported JSON data into MySQL database.
Here is the complete php code snippet I have used to insert JSON to MySQL using PHP.
<?php
//connecttomysqldb
$con=mysql_connect("username","password","")ordie('Couldnotconn
ect:'.mysql_error());
//connecttotheemployeedatabase
mysql_select_db("employee",$con);
//readthejsonfilecontents
$jsondata=file_get_contents('empdetails.json');
//convertjsonobjecttophpassociativearray
$data=json_decode($jsondata,true);
//gettheemployeedetails
$id=$data['empid'];
$name=$data['personal']['name'];
$gender=$data['personal']['gender'];
$age=$data['personal']['age'];
$streetaddress=$data['personal']['address']['streetaddress'];
$city=$data['personal']['address']['city'];
$state=$data['personal']['address']['state'];
$postalcode=$data['personal']['address']['postalcode'];
$designation=$data['profile']['designation'];
$department=$data['profile']['department'];
//insertintomysqltable
$sql="INSERTINTOtbl_emp(empid,empname,gender,age,streetaddres
s,city,state,postalcode,designation,department)
VALUES('$id','$name','$gender','$age','$streetaddress','$city',
'$state','$postalcode','$designation','$department')";
if(!mysql_query($sql,$con))
{
die('Error:'.mysql_error());
}
?>
Read:
How to Insert Multiple JSON Objects into MySQL in PHP
How to Convert MySQL to JSON Format using PHP
Hope this tutorial helps you to understand howto insert JSON data into MySQL using PHP.
Last Modified: Oct-11-2015

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Replies
Reply
44 comments:
Anonymous January 19, 2015 5:44 PM
How do you use this for more than one row of data? Say in the json there are two empid's,
and you want to insert each of them?
Reply
Valli Pandy January 19, 2015 11:30 PM
Hi, just loop through the json array if you have multiple rows like this.
//read the json file contents
$jsondata = file_get_contents('empdetails.json');
//convert json object to php associative array
$data = json_decode($jsondata, true);
foreach($data as $row)
{
//get the employee details
$id = $row['empid'];
...
//insert into db
mysql_query($sql,$con);
}
Umair January 31, 2015 7:42 PM
<?php
include 'include/connect.php';
error_reporting(E_ALL);
$jsondata = file_get_contents('D:xamphtdocslobstersapimonitoring_log.php');
if (is_array($data)) {
foreach ($data as $row) {

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$am = $row['item']['am'];
$pm = $row['item']['pm'];
$unit = $row['unit'];
$day = $row['day'];
$userid = $row['userid'];
$sql = "INSERT INTO monitoring_log (unit, am, pm, day, userid)
VALUES ('".$am."', '".$pm."', '".$unit."', '".$day."', '".$userid."')";
mysql_query($sql);
}
}
Reply
Umair February 01, 2015 1:02 AM
but sir still my code is not working. Can you solve my issue
Reply
Valli Pandy February 01, 2015 7:34 PM
Hi Umair, you haven't mentioned what error you are getting with the above code.
But I could see you are trying to run a "*.php" file and get the output. Passing the
exact file path to file_get_contents() won't execute the file.
Instead try using this,
file_get_contents('http://localhost/lobstersapi/monitoring_log.php')
Also make sure your web server is running for this to work.
Hope this helps you :)
Umair February 01, 2015 8:19 PM
still not working.
my json data that i am getting from iphone developer is {
"New item" : {
"PM" : false,
"AM" : false
},
"title" : " Unit 13"
}
and my code is :
<?php
include 'include/connect.php';
error_reporting(E_ALL);
$jsondata = file_get_contents('http://localhost/lobstersapi/monitoring_log.php');
if (is_array($data)) {

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Reply
$am = $row['New item']['am'];
$pm = $row['New item']['pm'];
$unit = $row['unit'];
$sql = "INSERT INTO monitoring_log (am, pm, unit)
VALUES ('".$am."', '".$pm."', '".$unit."')";
mysql_query($sql);
}
}
not showing any error.
Reply
Valli Pandy February 02, 2015 9:03 PM
Hi, I could see some inconsistencies in your code. Do you get the json output
properly? Even then, if the output contains single row like this,
{
"New item" : {
"PM" : "false",
"AM" : "false"
},
}
then you should not use foreach loop to parse it. Use the loop only if json output
contains multiple rows that looks something like this,
[{
"New item" : {
"PM" : "false",
"AM" : "false"
},
},
{
"New item" : {
"PM" : "false",
"AM" : "false"
},
}]
Also while retrieving data from json, you should use the exact key attribute like this,
$am = $row['New item']['AM'];
$pm = $row['New item']['PM'];
$unit = $row['title'];
(I could see you are using lower case and wrong key attributes here).
Note: If you don't get error print the values in browser and make sure you get the
desired result.

Replies
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Umair February 03, 2015 8:27 PM
Thank you so much sir
Reply
Anonymous February 27, 2015 2:47 PM
Nice tutorial,
But I want to store JSON data as a JSON into mysql, not the way you have done, then what
are the steps to store and retrieve JSON from mysql using php?
Reply
Valli Pandy February 28, 2015 12:57 AM
Hey! you can store json data in mysql without breaking up into columns. Only
proper escaping of json will protect the db from security risk.
But it's not a proper way to do as it just stores a long length of string and will
restrict mysql's ability to search, sort or index data and processing concurrent
queries.
This thread discusses the topic in more detail
http://stackoverflow.com/questions/20041835/putting-json-string-as-field-data-on-
mysql
Hope this helps :)
Vlad N April 03, 2015 12:43 PM
How can a json from an api be saved into mySQL? For example I have this json coming from
an api http://api.androidhive.info/json/movies.json. How can I save this data into the
database?
Reply
Valli Pandy April 04, 2015 1:22 AM
Hi,
If your json o/p is from remote server like the one you have mentioned
(http://api.androidhive.info/json/movies.json) then pass the complete url like this,
//read json file
$jsondata = file_get_contents('http://api.androidhive.info/json/movies.json');
This should work :)
JT May 25, 2015 4:17 AM
Hi,
I can't send my json data to mysql!! I need help...
$json_data = '[{
"external_urls" : "https://open.spotify.com/artist/2QWIScpFDNxmS6ZEMIUvgm",
"followers" : {

Replies
"total" : 116986
},
"genres" : "latin alternative",
"id" : "2QWIScpFDNxmS6ZEMIUvgm",
"name" : "Julieta Venegas",
"popularity" : 72,
"type" : "artist",
}]';
//convert to stdclass object
$data = json_decode($json_data, true);
$href = $data['external_urls'];
$followers = $data['followers']['total'];
$genres = $data['genres'];
$id = $data['id'];
$name = $data['name'];
$popularity = $data['popularity'];
$type = $data['type'];
//insert values into mysql database
$sql="INSERT INTO `spotify`(èxternal_urls`, `followers`, `genres`, èmpid`, èmpname`,
`popularity`, `type`)
VALUES ('$href', '$followers', '$genres', '$id', '$name', '$popularity', '$type')";
Reply
Valli Pandy May 25, 2015 5:21 AM
Hi, the json you have given is invalid with the comma (,) at the end of the last item
("type" : "artist",). It should be like,
$json_data = '[{
...
"popularity" : 72,
"type" : "artist"
}]';
Also the square brackets [] around json data makes it an array. So you should
iterate through it like this,
$href = $row['external_urls'];
$followers = $row['followers']['total'];
$genres = $row['genres'];
$id = $row['id'];
$name = $row['name'];
$popularity = $row['popularity'];
$type = $row['type'];
//insert values into mysql database
$sql="INSERT INTO `spotify`(èxternal_urls`, `followers`, `genres`, èmpid`,
èmpname`, `popularity`, `type`)
VALUES ('$href', '$followers', '$genres', '$id', '$name', '$popularity', '$type')";
}
This should work :)
JT May 25, 2015 5:49 AM
This work..!! :) Thank you so much Valli!!

Reply
Replies
Valli Pandy May 25, 2015 8:49 PM
Glad it works :)
igron July 13, 2015 6:48 PM
Thank you so much!
I'm new in web-programming and was feeling nervous with one of my first tasks. However
with your help i did a daytask within an hour.
Great!
Reply
Valli Pandy July 13, 2015 7:12 PM
Glad! I could help you...Cheers!!!
Gon July 30, 2015 6:22 AM
Hey valli, great work here, could you help me please?
I try to load a JSON named Business from this website
http://www.yelp.com/dataset_challenge. But give me an error on business_id, can t
load that.
my code is the following:
setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
echo "Connected successfully";
}
catch(PDOException $e)
{
echo "Connection failed: " . $e->getMessage();
}
$jsondata = file_get_contents('c:yelp_academic_dataset_business.json');
ini_set('memory_limit', '512M');
$idBusiness = $data['business_id'];
$neighborhoods = $data['neighborhoods'];
$full_address = $data['full_address'];
$city = $data['city'];
$state = $data['state'];
$latitude = $data['latitude'];
$longitude = $data['longitude'];
$stars = $data['stars'];
$review_count = $data['review_count'];
$open = $data['open'];
$procedure = $conn -> prepare("INSERT INTO business(business_id, name,
neighborhoods, full_address, city, state, latitude, longitude, stars, review_count,
open)
VALUES('$idBusiness', '$name', '$neighborhoods', '$full_address', '$city', '$state',
'$latitude', '$longitude', '$stars', '$review_count', '$open')");

$procedure -> execute(); ?>
Gon July 30, 2015 6:24 AM
Hey valli, great work here, could you help me please?
I try to load a JSON named Business from this website
http://www.yelp.com/dataset_challenge. But give me an error on business_id, can t
load that.
my code is the following:
setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
echo "Connected successfully";
}
catch(PDOException $e)
{
echo "Connection failed: " . $e->getMessage();
}
$jsondata = file_get_contents('c:yelp_academic_dataset_business.json');
ini_set('memory_limit', '512M');
$idBusiness = $data['business_id'];
$neighborhoods = $data['neighborhoods'];
$full_address = $data['full_address'];
$city = $data['city'];
$state = $data['state'];
$latitude = $data['latitude'];
$longitude = $data['longitude'];
$stars = $data['stars'];
$review_count = $data['review_count'];
$open = $data['open'];
$procedure = $conn -> prepare("INSERT INTO business(business_id, name,
neighborhoods, full_address, city, state, latitude, longitude, stars, review_count,
open)
VALUES('$idBusiness', '$name', '$neighborhoods', '$full_address', '$city', '$state',
'$latitude', '$longitude', '$stars', '$review_count', '$open')");
$procedure -> execute(); ?>
Gon July 30, 2015 6:24 AM
Could you help me? Then after make this work i need to load the field attributes to
a table in my sql named attributes to the fields Designation and value, How could i
do that, if there is so many attributes and i can t call them by your code, like
garage, parking, etc. take a look in the Json named business please.
Here's a example of a line of the JSON.
{"business_id": "fNGIbpazjTRdXgwRY_NIXA", "full_address": "1201 Washington
AvenCarnegie, PA 15106", "hours": {}, "open": true, "categories": ["Bars",
"American (Traditional)", "Nightlife", "Lounges", "Restaurants"], "city": "Carnegie",
"review_count": 5, "name": "Rocky's Lounge", "neighborhoods": [], "longitude":
-80.084941599999993, "state": "PA", "stars": 4.0, "latitude": 40.396468800000001,
"attributes": {"Alcohol": "full_bar", "Noise Level": "average", "Music": {"dj": false,
"background_music": true, "karaoke": false, "live": false, "video": false, "jukebox":
false}, "Attire": "casual", "Ambience": {"romantic": false, "intimate": false, "touristy":

false, "hipster": false, "divey": false, "classy": false, "trendy": false, "upscale": false,
"casual": false}, "Good for Kids": true, "Wheelchair Accessible": false, "Good For
Dancing": false, "Delivery": false, "Coat Check": false, "Smoking": "no", "Accepts
Credit Cards": true, "Take-out": false, "Price Range": 2, "Outdoor Seating": false,
"Takes Reservations": false, "Waiter Service": true, "Caters": false, "Good For":
{"dessert": false, "latenight": false, "lunch": false, "dinner": false, "brunch": false,
"breakfast": false}, "Parking": {"garage": false, "street": false, "validated": false, "lot":
false, "valet": false}, "Has TV": true, "Good For Groups": true}, "type": "business"}
Thank you :)
Valli Pandy July 31, 2015 11:54 PM
Hi, What error you get? From the json you have given below, the business id
seems to be a string value. Did you set the mysql 'datatype' for the said id field as
varchar or not? Please make sure you use the matching datatypes for the mysql
field attributes.
For the last query,
Your query is not clear and I hope this is what you want to ask.
The given json is a complex nested object and to get the value for the key 'garage'
you should use like this,
$garage = $data['attributes']['parking']['garage'];
If you have queries apart from this, please make it clear.
Cheers.
Anonymous August 13, 2015 7:56 AM
Is it possible to exclude certain lines from going into the database? My json file has
1 array with about 20 items in it. Do I just not write in those specific lines in the php
so it'll ignore them? I know the file is small, just doing this as a learning tool.
Valli Pandy August 15, 2015 3:12 AM
Hi, you can do it by checking on the key value on json records like this,
//convert json object to associative array
//loop through the array
if ($row['empid'] != '5121') { //replace this with your own filter
$id = $row['empid'];
....
//insert into mysql table
$sql = "INSERT INTO tbl_emp(empid, empname, gender, age, streetaddress, city,
state, postalcode, designation, department)
VALUES('$id', '$name', '$gender', '$age', '$streetaddress', '$city', '$state',
'$postalcode', '$designation', '$department')";
...
}
}
Hope this helps.
Cheers.

Reply
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KALYO August 24, 2015 3:41 PM
Great post, thanks ;-)
Reply
Valli Pandy August 25, 2015 1:29 AM
Glad you liked it :-)
paijo_jr September 11, 2015 1:21 PM
i will try thanks
Reply
BuxsCorner September 14, 2015 7:39 AM
Hi Vallie,
Hmm what if that was a dynamic JSon result, instead of INSERT syntax to MySQL, possible
to UPDATE syntax, or using PHP with if statement before INSERT-ing ?
Reply
Valli Pandy September 17, 2015 12:21 AM
Hi,
The process is same for dynamic json too, for e.g., you can make an api call and
store the result in a variable and proceed in the same way discussed above.
For your second question, yes you can use the json values with UPDATE query
also. As for using if statement, say you want to insert the values only for 'SALES'
department, then use it like
If ($department == "SALES") {
// Insert into DB
// or process the data as you wish
}
Hope that helps!
Cheers.
Rahul Jain September 15, 2015 9:58 PM
Hi,
I am trying to save the same json as mentioned above but not able to save data in mysql, the
code is working but it is showing empty fields in db.
and I have tried to echo the file contents and it is successful ,but the problem is that if I try to
echo the decoded data then it shows a blank page.
Reply

Replies
Reply
Valli Pandy September 17, 2015 12:29 AM
Hi! It seems like the json format you are trying is not valid. Please make sure you
haven't missed out quotes or any thing else.
You can check if you are using the right JSON format like this,
http://www.kodingmadesimple.com/2015/04/validate-json-string-php.html
Hope that helps.
Cheers.
Marjorie Lagos October 09, 2015 7:18 AM
Hello , how could I do with my code ?
I need to get the data :
Go dia_0 , dia_1 , dia_2 , dia_3 , dia_4 , dia_5 , dia_6 , dia_7 , DC , PMP and CR .
This is my code .
{ "type" : " FeatureCollection "
"features" : [
{ "Type" : " Feature "
"You properties" :
{
" Id" : "1 "
" dia_0 ": " 0 "
" dia_1 ": " 0 "
" dia_2 ": " 0 "
" dia_3 ": " 0 "
" dia_4 ": " 0 "
" dia_5 ": " 0 "
" dia_6 ": " 0 "
" dia_7 ": " 0 "
"CC" : "0 "
" PMP " , " 0 "
"CR ": " 0 "},
... please help
Reply
libraramis December 09, 2015 7:27 PM
Assalam-o-Alikum !!
i m trying to save data form json to sql by using following steps but i m getting error any help
??
//json file//
{
"error_code": "0",
"message": "success",
"food_items": [
{
"id": "1",
"food_name": "apple",
"food_fat": "10",
"food_bar_code": "25",
"food_carbs": "26",
"food_protein": "20",

Replies
"food_servings": "125",
"food_points": "2",
"food_fiber": "2"
}
]
}
// php code //
include 'config.php';
echo $jsondata = file_get_contents('document.json');
$food_name = $data['food_name'];
$food_fat = $data['food_fat'];
$food_bar_code = $data['food_bar_code'];
$food_carbs = $data['food_carbs'];
$food_protein = $data['food_protein'];
$food_servings = $data['food_servings'];
$food_points = $data['points'];
$food_fiber = $data['food_fiber'];
$addFood = "INSERT INTO food_points (food_name, food_fat, food_bar_code, food_carbs,
food_protein, food_servings, points, food_fiber)
VALUES
('$food_name', '$food_fat', '$food_bar_code', '$food_carbs', '$food_protein', '$food_servings',
'$food_points', '$food_fiber')";
if ($conn->query($addFood)===TRUE)
{
echo "data is updated . . . . !!!";
}
else
{
echo "Error: " . $addFood . "
" . $conn->error;;
}
// getting this error //
Notice: Undefined index: food_name in E:zhtdocsramistestindex.php on line 24
Notice: Undefined index: food_fat in E:zhtdocsramistestindex.php on line 25
Notice: Undefined index: food_bar_code in E:zhtdocsramistestindex.php on line 26
Notice: Undefined index: food_carbs in E:zhtdocsramistestindex.php on line 27
Notice: Undefined index: food_protein in E:zhtdocsramistestindex.php on line 28
Notice: Undefined index: food_servings in E:zhtdocsramistestindex.php on line 29
Notice: Undefined index: points in E:zhtdocsramistestindex.php on line 30
Notice: Undefined index: food_fiber in E:zhtdocsramistestindex.php on line 31
data is updated . . . . !!!
Reply

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Replies
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Valli Pandy December 11, 2015 1:38 AM
Welcome! Accessing your json data like this won't work. The key "food_items": [ ]
itself is representing an array. Note that the square brackets around [] is
considered as array and should be parsed with loop.
Working with json array is clearly described in another tutorial. Please check this
url: http://www.kodingmadesimple.com/2015/10/insert-multiple-json-data-into-
mysql-database-php.html
Let me know if you need any more help.
Cheers.
Unknown February 15, 2016 7:55 PM
Hi!
Thanx everybody for help, it's very useful!
How can I get and decode information from several json files in one php script? The files
have the simmilar structure. Could you transform the script for me?
Reply
Unknown February 15, 2016 7:55 PM
Hi!
Thanx everybody for help, it's very useful!
How can I get and decode information from several json files in one php script? The files
have the simmilar structure. Could you transform the script for me?
Reply
Valli Pandy February 19, 2016 4:52 AM
Hi! Say you have all the json files stored in a directory, you can read through the
files one by one like this.
$jdir = scandir($dirpath);
foreach($jdir as $file)
{
$jsondata = file_get_contents($file);
...
}
Cheers:)
Unknown February 25, 2016 12:40 AM
My error is-
PHP Notice: Undefined index: dishes in /Applications/MAMP/htdocs/ionicserver/start.php on
line 26

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Unknown March 05, 2016 10:03 PM
I am not able to view the data in phpmyadmin database
its showing data succefully inserted
my code is looks like this
Reply
Shahzad Ahmed March 22, 2016 7:08 AM
Thanks for share!
Reply
Valli Pandy March 26, 2016 3:33 AM
Welcome Ahmed:)
Unknown May 02, 2016 8:46 PM
hello,
I need help please: i have a php file which offer a jsonarray, i need to store this result into a
json file so i can explore it later !
For example for the command file_get_contents !
Any response please :( ?!
Reply
Chaima Bennour May 02, 2016 8:50 PM
Hi,
I need Help please !
I need to store the result provided by my php file (which is a jsonarray) into a json file, so i can
explore it later for example to use the command file_get_contents !
Any response please ?
Reply
Valli Pandy May 09, 2016 2:18 AM
We have an article that discusses the topic in detail. Please check this below link...
http://www.kodingmadesimple.com/2016/05/how-to-write-json-to-file-in-php.html
Cheers.

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