无法将多个json数据插入mysql数据库表

Question

I have this code.

<?php
// open mysql connection
$host = "localhost";
$username = "root";
$password = "";
$dbname = "jacklin";
$con = mysqli_connect($host, $username, $password, $dbname) or die('Error in Connecting: ' . mysqli_error($con));
// use prepare statement for insert query
$st = mysqli_prepare($con, 'INSERT INTO company_details(com_name, city, com_address, com_mno, com_lno, com_faxno, com_email, com_url, contact_person, com_img,
 lat, lng, cat_src_pos, state, country, password, status, plan, token, pin, contact_person1, contact_person2,
  com_mno1, com_mno2, fpass_token, adv_src_pos, alias, com_skype, cover)
VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)');
// bind variables to insert query params
mysqli_stmt_bind_param($st, 'ssssssssssssissssssssssssisss', $id, $city, $com_address, $com_mno, $com_lno, $com_faxno, $com_email, $com_url, $contact_person, $com_img, $lat, $lng, $cat_src_pos, $state, $country, $password, $status, $plan, $token, $pin, $contact_person1, $contact_person2, $com_mno1, $com_mno2, $fpass_token, $adv_src_pos, $alias, $com_skype, $cover);
// read json file
$filename = 'empdata.json';
$json = file_get_contents($filename);   
//convert json object to php associative array
$data = json_decode($json, true);
// loop through the array
foreach ($data as $row) {
    // get the employee details
    $id = $row['com_name'];
    $city = $row['city'];
    $com_address = $row['com_address'];
    $com_mno = $row['com_mno'];
    $com_lno = $row['com_lno'];
    $com_faxno = $row['com_faxno'];
    $com_email = $row['com_email'];
    $com_url = $row['com_url'];
    $contact_person = $row['contact_person'];
    $com_img = $row['com_img'];
    $lat = $row['lat'];
    $lng = $row['lng'];
    $cat_src_pos = $row['cat_src_pos'];
    $state = $row['state'];
    $country = $row['country'];
    $password = $row['password'];
    $status = $row['status'];
    $plan = $row['plan'];
    $token = $row['token'];
    $pin = $row['pin'];
    $contact_person1 = $row['contact_person1'];
    $contact_person2 = $row['contact_person2'];
    $com_mno1 = $row['com_mno1'];
    $com_mno2 = $row['com_mno2'];
    $fpass_token = $row['fpass_token'];
    $adv_src_pos = $row['adv_src_pos'];
    $alias = $row['alias'];
    $com_skype = $row['com_skype'];
    $cover = $row['cover']   
    // execute insert query
    mysqli_stmt_execute($st);
}
//close connection
mysqli_close($con);
?>

and my empdata.json is like.

[{"com_id":"1","com_name":"SORENTO GRANITO PVT.LTD","city":"Morbi","com_address":"8-A National High WayOld ghuntu Road ,Morbi - 363 642 (Guj.) INDIA","com_mno":"+919377721600","com_lno":"02822 - 243783 \/ 84","com_faxno":"(02822) 243785","com_email":"marketing@sorentogranito.com","com_url":"www.sorentogranito.com","contact_person":"Mr. Bhagubhai Tulsiyani","com_img":"1403952411.png","lat":"22.824254","lng":"70.8606801","cat_src_pos":"400000","state":"Gujarat","country":"India","password":"91SORESGPL","status":"active","plan":"premium","token":"","pin":"363 642","contact_person1":"","contact_person2":"","com_mno1":"","com_mno2":"","fpass_token":"","adv_src_pos":"400000","alias":"sorento-granito-pvt-ltd","com_skype":"","cover":"motto.jpg"},{"com_id":"3","com_name":"COTO CERAMIC PVT LTD","city":"Morbi","com_address":"8-A National Higway,B\/ h Makansar Panjarapore Weed...","com_mno":"+919099173713","com_lno":"+919099173713","com_faxno":"","com_email":"info@cotobathware.com","com_url":"www.cotobathware.com","contact_person":"Mr. SUMEET MARVANIYA","com_img":"d08687ba60bb3f0d1317e2fd8b10afd4.png","lat":"22.748123","lng":"70.9369573","cat_src_pos":"500000","state":"Gujarat","country":"India","password":"MAYANK8877","status":"active","plan":"basic","token":"","pin":"363621","contact_person1":"","contact_person2":"","com_mno1":"","com_mno2":"","fpass_token":"","adv_src_pos":"500000","alias":"coto-ceramic-pvt-ltd","com_skype":"","cover":"motto.jpg"},{"com_id":"4","com_name":"GLORY CERAMIC PVT LTD","city":"Morbi","com_address":"8\/A , National Highway Lalpar Morbi","com_mno":"+919825228848","com_lno":"02822 - 650445\/ 652446","com_faxno":"","com_email":"gloryceramic@yahoo.co.in","com_url":"www.gloryceramic.com","contact_person":"Mr. Niraj Thakkar","com_img":"1403952443.png","lat":"22.7968786","lng":"70.8907196","cat_src_pos":"80000","state":"Gujarat","country":"India","password":"9227650445","status":"active","plan":"premium","token":"","pin":"363 641","contact_person1":"","contact_person2":"","com_mno1":"","com_mno2":"","fpass_token":"","adv_src_pos":"80000","alias":"glory-ceramic-pvt-ltd","com_skype":"","cover":"motto.jpg"},{"com_id":"5","com_name":"SALON CERAMIC PVT.LTD.","city":"Morbi","com_address":"8-A National Highway, Olg Ghuntu Road, Morbi - 363 642(Guj.) INDIA","com_mno":"+91 9825223840","com_lno":"+91 2822 242115","com_faxno":"+91 2822 242116","com_email":"info@salonceramic.com","com_url":"www.salonceramic.com","contact_person":"Mr. Hiteshbhai","com_img":"1413397071.PNG","lat":"22.838649048614528","lng":"70.88279977525485","cat_src_pos":"400000","state":"Gujarat","country":"India","password":"123salon123","status":"active","plan":"premium","token":"252985240685b6f5b1728d0d31bc585b","pin":"363642","contact_person1":"","contact_person2":"","com_mno1":"","com_mno2":"","fpass_token":"","adv_src_pos":"400000","alias":"salon-ceramic-pvt-ltd","com_skype":"","cover":"motto.jpg"}]

with 1000 of records.but when i run above code in my localhost it's not displaying any error and not inserting any record to database too. please tell me how to insert this type json to database.

Answer 1

Try a simple foreach loop to count the number of data, and then make a for loop for running the insert statement:

$file = file_get_contents('empdata.json');
$json = json_decode($file, true);
//echo '<pre>';
//print_r($json);    
$num = array();//Open Blank array for number of data
foreach($json as $k => $v):
    $num [] = $v; //number of data        
endforeach;
$row= count($num);//Put number of data in $row
for($i=0; $i<$row; $i++){
   //instead of putting the value in the statement, store them in variable
   $com_name = $json[$i]['com_name'];
   $city = $json[$i]['city'];
   $com_address = $json[$i]['com_address'];
   //Put the other variable like this and put them in insert statement
   $stmt = mysqli_prepare($con, "INSERT INTO company_details VALUES (?, ?,?)");
   mysqli_stmt_bind_param($stmt, 'sss', $com_name, $city, $com_address);
}

This is just a simple working solution( working demo with real database ). I don't have no time for typing all the fields. For the second and third loop, there is no fax number, so if the data field default value cannot have null value, there will be some problem and your statement will not work properly and likely fails. I hope this will help.