Wie kaskadieren Sie drei Dropdown?Wie kaskadiere 3 Dropdown mit Javascript?
ich eine Seitenvorlage mit drei Drop-Down haben, sind die Werte der Option HTML-Tags dynamisch durch mysql füllen. Ich möchte die Optionen im 3. Dropdown-Menü anzeigen, hängt vom 2. Dropdown-Menü ab und zeigt die Optionen im 2. Dropdown-Menü an, abhängig von der im 1. Dropdown-Menü ausgewählten Option.
Ich habe versucht, einige JS-Skripte, aber ich habe kein Glück, es richtig zu arbeiten. Bitte helfen Sie mir, ich blieb fast 1 Woche auf diesem :(
PHP Snippet:
<?php
ob_start();
global $wpdb;
$query_position = "";
$list_position = "";
$result_position = "";
$query_locations = "";
$list_location = "";
$result_location = "";
$query_processed = "";
$list_processed = "";
$result_processed = "";
$query_position = $wpdb->get_results('SELECT DISTINCT position FROM resume_databank ORDER BY position ASC', OBJECT);
$query_locations = $wpdb->get_results('SELECT DISTINCT hiring_location FROM resume_location ORDER BY hiring_location ASC', OBJECT);
$query_processed = $wpdb->get_results('SELECT DISTINCT process_resume FROM resume_databank ORDER BY process_resume ASC', OBJECT);
?>
<div id="">
<form action='' method='post' name='resumeDatabank' id='resumeDatabank'>
<div class="div-select">
<label for="list_position" id="#ddress_search LABEL">Position</label>
<br/>
<select name="list_position" id="filterbypostion">
<option name="default" class="filter_by" selected="selected" value="Select by Position">Select by Position</option>
<?php
foreach($query_position as $option){
if(isset($_POST['list_position']) && $_POST['list_position'] == $option->position)
echo '<option name="list_position" class="filter_by" selected value="'. $option->position .'">'. $option->position .'</option>';
else
echo '<option name="list_position" class="filter_by" value="'. $option->position .'">'. $option->position .'</option>';
};
?>
</select>
</div>
<div class="div-select">
<label for="list_location" id="#ddress_search LABEL">Location</label>
<br/>
<select name="list_location" id="filterbylocation">
<option name="default" class="filter_by" selected="selected" value="Select by Location">Select by Location</option>
<?php
foreach($query_locations as $option){
if(isset($_POST['list_location']) && $_POST['list_location'] == $option->hiring_location)
echo '<option name="list_location" class="filter_by" selected value="'. $option->hiring_location .'">'. $option->hiring_location .'</option>';
else
echo '<option name="list_location" class="filter_by" value="'. $option->hiring_location.'">'. $option->hiring_location .'</option>';
};
?>
</select>
</div>
<div class="div-select">
<label for="list_processed" id="#ddress_search LABEL">Processed</label>
<br/>
<select name="list_processed" id="filterbyprocessed">
<option name="default" class="filter_by" selected="selected" value="Select by Processed">Select by Processed</option>
<?php
foreach($query_processed as $option){
if(isset($_POST['list_processed']) && $_POST['list_processed'] == $option->processed_option)
echo '<option name="list_processed" class="filter_by" selected value="'. $option->processed_option .'">'. $option->processed_option .'</option>';
else
echo '<option name="list_processed" class="filter_by" value="'. $option->processed_option.'">'. $option->processed_option .'</option>';
};
?>
</select>
</div>
<div class="div-input">
<input type="submit" value="Search" class="div-input-submit"/>
</div>
</form>
</div>
? was ist, wenn das js Skript und die php auf der gleichen Datei ist, wie es hier 'zu laden $ („# location“) Last („rpc.php“);.' – User014019
aber die js und pHP sind in der gleichen Seite? – User014019
. ich habe die Antwort-Code wird nun in derselben Seite – edd