Error while inserting a json array using php mysqli

Question

i have a array as json.stringify object. the array is from html table. i have made the array and it made well. i use the code below for create the array and try to getting the array for inserting into my database. i use ajax call function in there

$(document).on('click','#display_data',function(e){
    var convertTableToJson = function()
        {
            var rows = [];
            $('.table-bordered tr').each(function(i, n){
                var $row = $(n);
                rows.push([
                    $row.find('td:eq(0)').text(),
                    $row.find('td:eq(1)').text(),
                    $row.find('td:eq(2)').text(),
                    $row.find('td:eq(3)').text(),
                    $row.find('td:eq(4)').text(),
                    $row.find('td:eq(5)').text(),
                    $row.find('td:eq(6)').text(),
                    $row.find('td:eq(7)').text(),
                ]);
            });
            return JSON.stringify(rows);
        };
	$.ajax({
			data:convertTableToJson,
			type:"POST",
			url:"../php/tagihan/save_data.php",
			success: function(data){
				alert ("Data:" + data);
			}
		});
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.6.3/jquery.min.js"></script>
<table class="table table-bordered">
<thead>
<tr>
<th>Kode Material</th>
<th>Storage Location</th>
<th>Movement Type</th>
<th>Id Indentifier</th>
<th>Item</th>
<th>Date Input</th>
<th>Netto</th>
<th>Unit</th>
<th><input type="checkbox" id="mycheckbox1" name="mycheckbox1"></th>
</tr>
</thead>
<tbody>
<tr>
<td>101200</td>
<td>WCB</td>
<td>101</td>
<td>5006540050</td>
<td>1</td>
<td>10.08.2017</td>
<td>23.970</td>
<td>KG</td>
<td><input type="checkbox" id="mycheckbox" name="mycheckbox"></td>
</tr>
<tr>
<td>101200</td>
<td>WCB</td>
<td>101</td>
<td>5006539985</td>
<td>1</td>
<td>10.08.2017</td>
<td>42.970</td>
<td>KG</td>
<td><input type="checkbox" id="mycheckbox" name="mycheckbox"></td>
</tr>
</tbody>
</table>
<input type="button" id="display_data" name="display_data" />

i don't know why. when i try to run this code with this php function i get an error say's invalid argument of foreach(). and i think the syntax of the foreach that i use is right the php function that i use for insert the data just like the code below

<?php
include('../../Connections/koneksi.php');

// reading json file
$array = json_decode($_POST['convertTableToJson']);
$data = json_decode($array, true);
foreach ($data as $user) {
    $kode = $user[0];
    $sloc = $user[1];
    $move = $user[2];
    $id = $user[3];
    $date = $user[4];
    $netto = $user[5];
    $unit = $user[6];
    $payroll = $user[7];

    // preparing statement for insert query
    $st = mysqli_prepare($db, 'INSERT INTO wjm(kode, sloc, move, id, date, netto, unit, payroll) VALUES (?, ?, ?, ?, ?, ?, ?, ?)');

    // bind variables to insert query params
    mysqli_stmt_bind_param($st, 'ssssssss', $kode, $sloc, $move, $id, $date, $netto, $unit, $payroll);

    // executing insert query
    mysqli_stmt_execute($st);
}

?>

please someone help me to solve this.

You're using json_decode twice; this might be your problem

ahmad
– ahmad

2017-08-19 13:32:24 +00:00
Commented Aug 19, 2017 at 13:32 — ahmad
– ahmad, Commented Aug 19, 2017 at 13:32

Raul Valverde · Accepted Answer · 2017-08-19 13:34:09Z

1

If you want to obtain the values of a foreach it is necessary to obtain in foreach the key value

foreach ($data as $key => $value) {
    $kode       = $value[0];
    $sloc       = $value[1];
    $move       = $value[2];
    $id         = $value[3];
    $date       = $value[4];
    $netto      = $value[5];
    $unit       = $value[6];
    $payroll    = $value[7];

    ...
}

More info: http://php.net/manual/en/control-structures.foreach.php

answered Aug 19, 2017 at 13:34

Raul Valverde

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Minervaz Mine · Accepted Answer · 2017-08-19 13:39:09Z

0

did you make sure $array = json_decode($_POST['convertTableToJson']); getting the array object?

answered Aug 19, 2017 at 13:39

Minervaz Mine

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1 Comment

YvesLeBorg Over a year ago

use comments to requests clarifications. answers are for ... well answers.

surge10 · Accepted Answer · 2017-08-19 18:09:47Z

0

After you fix the syntax error with @Raul code, you will run into pragmatic error. Is should be writing with an if/else statement inside a for loop. so what you need is a for loop to avoid duplicates as in the case of @Raul code.

for ($x = 0; $x < count($data); $x++) {
    if($x == 0) {
        $kode = $data[0];
    } else if($x == 1) {
        $sloc = $data[1];
    } else if($x == 2) {
        $move = $data[2];
    } else if($x == 3) {
        $id = $data[3];
    } else if($x == 4) {
        $date = $data[4];
    } else if($x == 5) {
        $netto = $data[5];
    } else if($x == 6) {
        $unit = $data[6];
    } else if($x == 7) {
        $payroll = $data[7];
    } else { // do nothing }
}

answered Aug 19, 2017 at 18:09

surge10

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