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PHP&AJAX填充选择列表_Php_Jquery_Ajax - Fatal编程技术网

PHP&AJAX填充选择列表

PHP&AJAX填充选择列表,php,jquery,ajax,Php,Jquery,Ajax,我对填充选择框有疑问。 我有第一个选择框id='select_proizvodjaci',它工作正常,每次更改时,我都从函数$modelsauto=$proizvodjac->pretragamodelaproizvodjacu$pretraga中获得了正确的数据;在萤火虫中。但我不知道如何在proizvodjac.php中填充选择框select id='searchModel' 这是我的密码: proizvodjac.php <form method='post' action='pro

我对填充选择框有疑问。 我有第一个选择框id='select_proizvodjaci',它工作正常,每次更改时,我都从函数$modelsauto=$proizvodjac->pretragamodelaproizvodjacu$pretraga中获得了正确的数据;在萤火虫中。但我不知道如何在proizvodjac.php中填充选择框select id='searchModel'

这是我的密码: proizvodjac.php

<form method='post' action='proizvodjac-select.php' id='proizvodjacForma'>
        <select name='pretraga' id='select_proizvodjaci'>
            <option value=''>Izaberi</option>
            <?php foreach($proizvodjaci as $sets): ?>
                    <option value='<?php echo $sets['id']; ?>'><?php echo $sets['naziv_proizvodjaca']; ?></option>
            <?php endforeach; ?>
        </select>
    </form>

<label>Model:</label><br/><br/>
    <select id='searchModel'>
        <?php foreach($modelAuta as $jedanModel): ?>
        <option value='<?php echo $jedanModel['id']; ?>'><?php echo $jedanModel['naziv_modela']; ?></option>
    <?php endforeach; ?>
    </select>
proizvodjac-select.php

<?php 

include 'ProizvodjacModel.php';

// 1. Create database connection
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'radni_nalog_oop2';
$connection = new mysqli($dbhost, $dbuser, $dbpass, $dbname);

// Test if connection occurred.
if (mysqli_connect_errno()) {
    die('Database connection failed: ' .
        mysqli_connect_error() . 
        ' (' . mysqli_connect_errno() . ')'
    );
} else {
    echo 'Connection is successfully!' . '<br/>';
}

$proizvodjac = new ProizvodjacModel($connection);


$proizvodjaci = $proizvodjac->dajProizvodjace();

if(isset($_POST['pretraga'])) {

    $pretraga = $_POST['pretraga'];

    $modelAuta = $proizvodjac->pretragaModelaPoProizvodjacu($pretraga);


    print_r($modelAuta);

    $model_option = '<option>---</option>';


}

?>
如果你给

数据类型:“json”

ajax返回类型应为json格式,否则将无法工作

删除该行或使用

数据类型:“html”->如果您需要html
数据类型:'text'->如果您需要纯文本$searchModel.htmldata{SELECT BOX KEY};谢谢你的帮助,但是什么是数据{selectbox KEY}?实际上是数据['selectboxHTML'];这个键应该包含来自ajaxI found解决方案的HTML,我删除了ajax中的数据类型,并将javascript更改为:$.ajax{url:'proizvodjac select.php',//数据类型:'json',类型:'post',数据:'pretraga='+$'select_proizvodjaci.val}.donefunctiondata{$'searchModel.htmldata;};
<?php 

include 'ProizvodjacModel.php';

// 1. Create database connection
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'radni_nalog_oop2';
$connection = new mysqli($dbhost, $dbuser, $dbpass, $dbname);

// Test if connection occurred.
if (mysqli_connect_errno()) {
    die('Database connection failed: ' .
        mysqli_connect_error() . 
        ' (' . mysqli_connect_errno() . ')'
    );
} else {
    echo 'Connection is successfully!' . '<br/>';
}

$proizvodjac = new ProizvodjacModel($connection);


$proizvodjaci = $proizvodjac->dajProizvodjace();

if(isset($_POST['pretraga'])) {

    $pretraga = $_POST['pretraga'];

    $modelAuta = $proizvodjac->pretragaModelaPoProizvodjacu($pretraga);


    print_r($modelAuta);

    $model_option = '<option>---</option>';


}

?>