[英]jQuery function doesn't return anything on callback
我有一個jQuery函數,它采用表單select元素的selected選項並將其發送到php文件,該文件應該返回html來填充第二個select元素。 不幸的是,jQuery正在觸發,但是返回為空。 有任何想法嗎?
表單元素:
if ($result) {
echo '<label>*Team: <select name="team" class="team" style=\'width: 150; font-size: 16px;\' autocomplete="off" tabindex="1">';
echo '<option value="">Select</option>';
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo '<option value="'.$row['TeamID'].'">'.$row['CaptainLast']." ".date('m-y',strtotime($row['ArrDate'])).'</option>';
}
echo '</select></label>';
} else {
echo "0 results";
}
?>
<label>*Team Member: <select name="member" class="member" style='width: 150; font-size: 16px;' autocomplete="off" tabindex="2">
<option value="">Select</option>
</select></label>
jQuery:
$(document).ready(function(){
$(document).on('change','.team', function(){
var id=$(".team option:selected").val();
var dataString = 'tmid='+ id;
console.log(dataString);
$.ajax({
type: "POST",
url: "scripts/memfltpop.php",
data: dataString,
dataType: 'html',
success: function(html){
$(".member").html(html);
}
});
});
});
和服務器文件:
<?php
$team = $_POST['tmid'];
echo $team;
$con = mysqli_connect('**********', '****', '******', '*******');
if ($con) {
$sql = "SELECT MemberAdmin.IndID, MemberAdmin.First, MemberAdmin.Last ". "FROM MemberAdmin ". "LEFT JOIN Members ON MemberAdmin.IndID = Members.IndID ". "WHERE Members.TeamID='" .$team. "'";
$result = mysqli_query($con,$sql) or die(mysqli_error());
if ($result) {
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$id=$row['IndID'];
$data=$row['First']." ".$row['Last'];
echo '<option value="">Select</option>';
echo '<option value="'.$id.'">'.$data.'</option>';
}
} else {
echo "0 results";
}
}
else {
echo'Not connected to database';
}
為了獲得html中的響應,您需要在ajax請求中使用datatype html,如下所示:
$.ajax({
type: "POST",
url: "scripts/memfltpop.php",
data: dataString,
dataType : 'html',
success: function(html){
$(".member").html(html);
}
});
哇,找到了 我從沒想過這會導致SQL崩潰,但是我從
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
至
while($row = mysqli_fetch_array($result, MYSQL_ASSOC)) {
在更新SQL語言時,我錯過了這一行。 多謝你們!
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