Js，Ajax，Json - 如何不从自动完成中将“数组”插入到数据库中

Question

im building a little script that will automatically fill few input fields based on autocomplete of the first input field.

I have the script working, but when i hit submit button everything seems to be ok, But when i look into the db to see what was inserted, All fields are inserted with "Array" instead of what supposed to be inserted. on the autocomplete, I can see that field are being field with the right info, But i guess php doesn't understand what it is, and inserts "Array" instead.

Any ideas how to fix it?

my form:

<form action="job_post.php" method="post">
                    <div class="form-group has-success col-md-3">
                    <label class="control-label" for="inputSuccess1">First Name: </label>
                    <input type='text' id='firstname' name='firstname[]'/>
                </div>
                <div class="form-group has-success col-md-3">
                    <label class="control-label" for="inputSuccess1">Last Name: </label>
                    <input type='text' id='lastname' name='lastname[]'/>
                </div>
                <div class="form-group has-success col-md-3">
                    <label class="control-label" for="inputSuccess1">Age: </label>
                    <input type='text' id='age' name='age[]'/>
                </div>
                <div class="form-group has-success col-md-3">
                    <label class="control-label" for="inputSuccess1">Company Name: </label>
                    <input type='text' id='CoName' name='CoName[]'/>
                </div>

<input type="submit">
</form>

My job_post.php

<?php

require_once 'db_connect.php';

$firstname=$_POST['firstname'];
$lastname=$_POST['lastname'];
$age=$_POST['age'];
$CoName=$_POST['CoName'];

$qry=mysql_query("INSERT INTO jobs (FirstName, LastName, Age, CoName)
VALUES ('$firstname', '$lastname', '$age', '$CoName')", $con);
if(!$qry)
{
mysql_close($con);
die("Query Failed: ". mysql_error());
}

else
{
mysql_close($con);
header("Location:index.php");
}

mysqli_close($con);
?>

My Js:

$('#firstname').autocomplete({
                    source: function( request, response ) {
                        $.ajax({
                            url : 'ajax.php',
                            dataType: "json",
                            data: {
                               name_startsWith: request.term,
                               type: 'firstname',
                               row_num : 1
                            },
                             success: function( data ) {
                                 response( $.map( data, function( item ) {
                                    var code = item.split("|");
                                    return {
                                        label: code[0],
                                        value: code[0],
                                        data : item
                                    }
                                }));
                            }
                        });
                    },
                    autoFocus: true,            
                    minLength: 2,
                    select: function( event, ui ) {
                        var names = ui.item.data.split("|");
                        console.log(names[1], names[2], names[3]);                      
                        $('#lastname').val(names[1]);
                        $('#age').val(names[2]);
                        $('#CoName').val(names[3]);
                    }               
                  });

My Ajax:

<?php
/*
Site : http:www.smarttutorials.net
Author :muni
*/
require_once 'db_connect.php';

if($_GET['type'] == 'firstname'){
    $row_num = $_GET['row_num'];
    $result = mysql_query("SELECT FirstName, LastName, Age, CoName FROM jobs where FirstName LIKE '".strtoupper($_GET['name_startsWith'])."%'");    
    $data = array();
    while ($row = mysql_fetch_array($result)) {
        $name = $row['FirstName'].'|'.$row['LastName'].'|'.$row['Age'].'|'.$row['CoName'].'|'.$row_num;
        array_push($data, $name);   
    }   
    echo json_encode($data);
}


?>

It does pull data, but im looking to insert it to the db with submit button.

Any help highly appreciated.

Answer 1

Updated info

Then place the variables in the query you have now.

Page redone:

 <?php

    require_once 'db_connect.php';

    $firstname = $_POST["firstname"][0];
    $CoName = $_POST["CoName"][0]; 

    /*
     * Last Name and Age are not in the array, I removed them from the query 
     * however you need to place them in the array if you wish to add them
     */

    $qry=mysql_query("INSERT INTO jobs (FirstName, CoName)
        VALUES ('$firstname', '$CoName')", $con);

    if(!$qry) {
        mysql_close($con);
        die("Query Failed: ". mysql_error());
    } else {
        mysql_close($con);
        header("Location: index.php");
    }

    ?>

Updated again, misread array.