我對AJAX請求有麻煩,我不知道爲什麼。以下代碼似乎發送到整個網頁腳本(如我的警報框和控制檯中所示),而不是我的複選框值。任何人都可以向我解釋我做錯了什麼? 這是我的PHP複選框,具有由SQL生成的值,並沒有提交按鈕,使代碼設置爲從用戶改變運行:AJAX似乎是發送整個網頁腳本而不是數據
<form id="numberOrderForm" action="testdatabase.php" method="post">
<div class="wrappers" id="multi-select1Wrapper">
<h2>Area Code</h2>
<select class="dropDownMenus" id="multi-select1" name="multi_select1[]" multiple="multiple">
<?php
//The query asking from our database
$areaCodeSQL = "SELECT ac.Number AS `AreaCode`, ac.Name AS `AreaName`
FROM `AreaCodes` ac"; //SQL query: From the table 'AreaCodes' select 'Number' and put into 'AreaCode', select Name and put into 'AreaName'
$areaCodeResults = $conn->query($areaCodeSQL); // put results of SQL query into this variable
if ($areaCodeResults->num_rows > 0) { // if num_rows(from $results) is greater than 0, then do this:
// output data of each row
foreach($areaCodeResults as $areaCodeResult) //for each item in $areCodeResults do this:
{
$areaNameAndCode = $areaCodeResult['AreaCode'] ." ". $areaCodeResult['AreaName']; //get AreaCode and AreaName from query result and concat them
$areaName = $areaCodeResult['AreaName']; // get AreaName
$areaCode = $areaCodeResult['AreaCode']; //get AreaCode
?><option class="menuoption1" name="menuAreaCode" value="<?php echo $areaCode ?>" ><?php echo $areaNameAndCode; ?></option><?php //Create this option element populated with query result variables
}
}
?>
</select>
</div>
</form>
這裏是我的jQuery AJAX代碼:
<script>
$('#multi-select1').on("change", function(){
var areaCode = $(this).val();
$.ajax({
type:"POST",
url: "testdatabase.php", //my PHP database
data: "areacode=" + areaCode,
success: function(response){
//do stuff after the AJAX calls successfully completes
alert (response);
console.log(response);
},
error : function(xhr, status, error) {
alert(xhr.responseText);
}
});
});
</script>