Php 从选项中获取值
你好,我正在做一些尝试和错误。这是select选项从数据库填充的代码,但这给了我空值Php 从选项中获取值,php,jquery,ajax,Php,Jquery,Ajax,你好,我正在做一些尝试和错误。这是select选项从数据库填充的代码,但这给了我空值 echo "<option value=\"\">"."Select"."</option>"; $qry = "select * from try where name = '".$_POST['name']."'"; $result = mysqli_query($con,$qry); while ($row = mysqli_fetch_array($result)){
echo "<option value=\"\">"."Select"."</option>";
$qry = "select * from try where name = '".$_POST['name']."'";
$result = mysqli_query($con,$qry);
while ($row = mysqli_fetch_array($result)){
echo "<option value='".$row['trynum']."'>".$row['tryname']."</option>";
}
我的问题是我是否从下拉列表中选择内容,但没有价值。请帮忙
data: {instructor:$('#SELECT_ELEMTN_ID').val()},
根据范围和内容的不同,您可能不想使用它。PHP:
Jquery
phpfile.php
您的数据类型为“json”,但返回的是html.Protip:this.value>$this.val如果您在查看AJAX请求/响应时打开浏览器上的控制台窗口,您会看到@JasonP所说的话。顺便说一句,您很容易受到SQL注入的攻击。快速回复。。我应该使用什么数据类型?我可以在哪里阅读任何链接?
<select id="sub" name="subb"></select>
data: {instructor:$('#SELECT_ELEMTN_ID').val()},
$ajaxAnswer = "<option value=\"\">"."Select"."</option>";
$instructor = mysqli_real_escape_string($conn,$_POST['instructor']);
$qry = "select * from try where name = '".$instructor."'";
$result = mysqli_query($con,$qry);
while ($row = mysqli_fetch_array($result)){
$ajaxAnswer .= "<option value='".$row['trynum']."'>".$row['tryname']."</option>";
}
echo $ajaxAnswer;
$.ajax({
type: "POST",
url: "json_php_sub.php",
data: {instructor:$(this).val()},
success: function(result){
$("#sub").html(result);
}
});
$(document).ready(function () {
$.ajax({
type: "GET",
url: "phpfile.php",
dataType: "json",
success: function (data) {
$.each(data, function (idx, obj) {
$('#selectdata').append('<option value="'+obj.user_id+'">'+obj.user_name+'</option>' )
});
}
});
});
</script>
</head>
<body>
<select id="selectdata">
</select>
</body>
<?php
$host = "localhost";
$user = "root";
$password ="";
$database= "databasename";
$con = mysqli_connect($host , $user , $password);
$database_connect = mysqli_select_db($con, $database);
$result = mysqli_query($con, "select Id as user_id,Name as user_name from users");
$data = mysqli_fetch_all($result, MYSQLI_ASSOC);
echo json_encode($data);
?>