2016-10-17 34 views
0

您好,我試圖將多個選定的值存儲在一個新的表中最後插入的ID。如何存儲多個複選框,如果它們是動態的在php

我的錯誤是數組來了,但id_facility值不來

這裏是我的形式:

<ul id="hostel_facility" class="dropdown-menu dropdown-select"> 
    <?php $facility = $conn->query("SELECT * FROM os_facilities ORDER BY id_facility ASC"); 
     while ($facilityresult = $facility->fetch_assoc()) { ?> 
     <li><a><input type="checkbox" name="hostel_facility[]" value="<?php echo$facilityresult['id_facility']; ?>" /><?php echo $facilityresult['facility_name']; ?></a></li> 
    <?php } ?> 
</ul> 

我想在新表中的列存儲的id_hostel,id_facility

+1

我檢查你的代碼,它的工作。我檢查'inspect'元素名稱應該是'hostel_facility []'值'1'到'n'。你的pblm是什麼? – Karthi

+0

已解決.. –

+0

你有什麼錯誤? – Karthi

回答

0

我的形式:

<ul id="hostel_facility" class="dropdown-menu dropdown-select"> 
    <?php $facility = $conn->query("SELECT * FROM os_facilities ORDER BY id_facility ASC"); 
     while ($facilityresult = $facility->fetch_assoc()) { ?> 
     <li><a><input type="checkbox" name="hostel_facility[]" value="<?php echo $facilityresult['id_facility']; ?>" /><?php echo $facilityresult['facility_name']; ?></a></li> 
    <?php } ?> 
</ul> 

這裏是存儲作業後,我的控制器頁:

$hostelId=$conn->insert_id; 
if($addHostel) 
{ 
foreach($hostelData['hostel_facility'] as $key=>$facility) 
{ 
    $conn->query("INSERT INTO os_hostel_facility (id_hostel,id_facility) values ('".$hostelId."','".$facility."')"); 
} 
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