我有一個用PHP和MYSQL編寫的網站,其中包含3個獨立drodpdown列表,其中第一個下拉列表包含了其他2個dorpdown列表所必需的多個值。如何在第二個變量裏面存儲下拉列表中的值
我需要的是也能夠將這些值之一存儲在第二個變量中,以便在其他查詢中單獨使用它。
代碼
<?php
// code for submit button action
global $wpdb, $site_name, $data;
//variables that handle the retrieved data from mysql database based on the ID of the variable in HTML (select)
if(isset($_POST['query_submit']))
{
if(isset($_POST['site_name']))
{
$site_name=$_POST['site_name'];
}
else { $site_name=""; }
var_dump($site_name);
/*在上述行添加
後續代碼var_dump顯示器3倍$ SITE_NAME變量內的值是有辦法來修整或SQL查詢之前提取第三值下。 */
$sql = $wpdb->prepare("select i.siteID
, i.siteNAME
, i.equipmentTYPE
, c.latitude
, c.longitude
, c.height
, o.ownerNAME
, o.ownerCONTACT
, x.companyNAME
, y.subcontractorCOMPANY
, y.subcontractorNAME
, y.subcontractorCONTACT
from site_info i
LEFT
JOIN owner_info o
on i.ownerID = o.ownerID
LEFT
JOIN company_info x
on i.companyID = x.companyID
LEFT
JOIN subcontractor_info y
on i.subcontractorID = y.subcontractorID
LEFT JOIN site_coordinates2 c
on i.siteID=c.siteID
where
i.siteNAME = %s
AND
o.ownerNAME = %s
AND
x.companyNAME = %s
",$site_name,$owner_name,$company_name);
$query_submit =$wpdb->get_results($sql, OBJECT);
代碼
<td><select id="site_name" name = "site_name">
<option value="">Select Site</option>
<?php
$query_site_name =$wpdb->get_results("select DISTINCT
i.siteNAME,
i.ownerID,
i.companyID,
o.ownerNAME,
x.companyNAME
from site_info i
LEFT
JOIN owner_info o
on i.ownerID = o.ownerID
LEFT
JOIN company_info x
on i.companyID = x.companyID
");
foreach($query_site_name as $row)
{
//$result1 = $row->ownerID;
// $result2 = $row->companyID;
echo "<option value = '".$row ->ownerID.",".$row ->companyID.",".$row ->siteNAME."'>".$row->siteNAME."</option>";
// echo "<option value = '".$row ->siteNAME."'>".$row->siteNAME."</option>";
}
?>
</select></td>
Ajax代碼
<script type="text/javascript">
// make Dropdownlist depend on each other
$(document).ready(function(){
// depend owner name on site name
$('#site_name').change(function(){
var arrayId = $(this).val().split(",");
if(arrayId != ""){
var ownerID = arrayId[0]; //0
var companyID = arrayId[1]; //1
$.ajax({
url:"<?php echo get_stylesheet_directory_uri(); ?>/dropdown_fetch_owner.php",
method:"POST",
data:{ownerID:ownerID,companyID:companyID},
dataType:"text",
success:function(data){
var Response = data.split("--");
$('#owner_name').html(Response[2]);
$('#Company_name').html(Response[4]);
}
});
}
});
});
</script>
dropdown_fetch_owner.php
<?php
include_once($_SERVER['DOCUMENT_ROOT'].'/wordpress/wp-load.php');
global $wpdb,$owner_name,$company_name;
$sql =$wpdb->get_results("select ownerID, ownerNAME from owner_info where ownerID = '".$_POST['ownerID']."' ORDER BY ownerNAME");
$owner_name = '--Owner--';
var_dump($sql);
foreach($sql as $row){
$owner_name.= "<option value ='".$row ->ownerID."'>".$row->ownerNAME."</option>";
}
echo $owner_name;
$sql =$wpdb->get_results("select companyID, companyNAME from company_info where companyID = '".$_POST['companyID']."' ORDER BY companyNAME");
$company_name = '--Company--';
var_dump($sql);
foreach($sql as $row){
$company_name.= "<option value ='".$row ->companyID."'>".$row->companyNAME."</option>";
}
echo $company_name;
exit();
?>
您需要存儲之後的第二信息從第一個下拉列表中選擇一個選項? –
@HamzaAbdaoui我需要將** siteNAME **值存儲在一個單獨的變量中,或將其他2個值存儲在一個單獨的變量中,以便能夠在第二個查詢中只使用siteNAME ..我將編輯我的問題並添加一些代碼看看 –
不是'var arrayId = $(this).val()。split(「,」); if(arrayId!=「」){var ownerID = arrayId [0]; // 0 var companyID = arrayId [1]; // 1'做這份工作? –