jQuery JSON下拉列表不填充 [英] jQuery JSON drop down not populating
问题描述
我正在尝试使用jQuery,PHP和mySQL创建一个下拉列表,这是到目前为止的代码,
I am trying to create a drop down list using jQuery, PHP and mySQL, here is my code thus far,
HTML:
<select id="box"></select><br />
<input id="button" type="button" value="populate" />
Javascript/jQuery:
Javascript/jQuery:
$(document).ready(function() {
$("#button").click(function (){
$.getJSON("test_post.php", function(data){
$.each(data, function(user) {
$('#box').append(
$('<option></option>').html(user)
);
});
});
});
});
PHP:
mysql_connect('localhost', 'user', '1234');
mysql_select_db('json');
$test = mysql_query('SELECT user,pass FROM login WHERE user="john"');
while($row = mysql_fetch_array($test, true)) {
$data .= json_encode($row);
};
echo $data;
我在数据库中有3个条目,当我运行'test_post.php'文件时,它会回显JSON,但是当我尝试获取它以填充下拉列表时,什么都没有发生!
I have 3 entries in the database, and when I run the 'test_post.php' file, it echoes out the JSON, but when I try and get it to populate the drop down, nothing happens!
知道为什么吗?
推荐答案
您的代码中存在几个问题.
There are several problems in your code.
首先::尝试使用未定义的变量$data
,您应该在while
循环:$data = '';
之前将其初始化为空字符串,否则它可以通过JSON响应发送PHP通知,具体取决于display_errors
和error_reporting
设置的值.
First: attempt to use undefined variable $data
, you should have initialized it to an empty string before while
loop: $data = '';
, otherwise this can send PHP notice with the JSON response, depending on values of display_errors
and error_reporting
settings.
第二个:如@shamittomar所说,$data
必须是一个数组,而json_encode()
对于整个数组只能被调用一次.目前,您正在发送几个串联到单个字符串的JSON对象,这是错误的.
Second: as @shamittomar said, $data
must be an array and json_encode()
must be called only once for the whole array. For now, you're sending several JSON objects concatenated to a single string, which is wrong.
第三条:JavaScript中user
函数参数的值实际上是data
数组的 index ,而不是值,但即使user
是值,也将是具有user
和pass
属性的JavaScript 对象,而不是字符串.
Third: value of the user
function parameter in the JavaScript is actually an index of the data
array, not a value, but even if user
would be a value, it would be a JavaScript object with user
and pass
properties, not a string.
结果代码应为(仅更改部分):
Resulting code should be (changed parts only):
PHP
$data = array();
while ($row = mysql_fetch_array($test, true)) {
$data[] = $row;
};
echo json_encode($data);
JavaScript
$.each(data, function(i, user) {
$('#box').append(
$('<option></option>').html(user.user)
);
});
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