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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> 

Antwort

0

Hier ist meine Probe (adaptieren es für Ihr Beispiel), ich hoffe, zu helfen:

<? 
mysql_connect('localhost','root',''); 
mysql_select_db('test'); 
if(!empty($_POST['location'])) { 
$qry = mysql_query("SELECT * FROM location"); 
echo '<select id="location">'; 
while($rw = mysql_fetch_row($qry)) { 
echo '<option value="'.$rw[1].'">'.$rw[1].'</option>'; 
} 
echo '</select>'; 
} else { 
$query = mysql_query("SELECT * FROM job"); 
echo '<select id="job">'; 
while($row = mysql_fetch_row($query)) { 
echo '<option value="'.$row[0].'">'.$row[1].'</option>'; 
} 
echo '</select>'; 
} 
?> 
<script type="text/javascript" src="jquery.js"></script> 
<script type="text/javascript"> 
$(document).ready(function(){ 
$("#job").bind("change", function(){ 
$.post("index.php",{'location': $('#job').val()},function(data){ 
$('#location').remove(); 
$("#job").after(data); 
}); 
}); 
}); 
</script> 
+0

? was ist, wenn das js Skript und die php auf der gleichen Datei ist, wie es hier 'zu laden $ („# location“) Last („rpc.php“);.' – User014019

+0

aber die js und pHP sind in der gleichen Seite? – User014019

+0

. ich habe die Antwort-Code wird nun in derselben Seite – edd