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我想通过使用select2插件将MySQL远程数据用作标记。但是我的JSON学习曲线非常薄弱。请帮我修正这个JSON格式。这个JSON为什么不起作用?
$("#e7").select2({
placeholder: "Search for a movie",
minimumInputLength: 3,
ajax: {
url: "search.php",
dataType: 'jsonp',
quietMillis: 100,
data: function (term, page) { // page is the one-based page number tracked by Select2
return {
q: term, //search term
page_limit: 10, // page size
page: page, // page number
apikey: "ju6z9mjyajq2djue3gbvv26t" // please do not use so this example keeps working
};
},
results: function (data, page) {
var more = (page * 10) < data.total; // whether or not there are more results available
// notice we return the value of more so Select2 knows if more results can be loaded
return {
results: data.movies,
more: more
};
}
},
formatResult: movieFormatResult, // omitted for brevity, see the source of this page
formatSelection: movieFormatSelection, // omitted for brevity, see the source of this page
dropdownCssClass: "bigdrop", // apply css that makes the dropdown taller
escapeMarkup: function (m) {
return m;
} // we do not want to escape markup since we are displaying html in results
});
这里是search.php中
<?php
$sql=mysqli_query($db3->connection,"SELECT * FROM tags");
while($row=mysqli_fetch_array($sql)){
$tags=$row['tags'];
$id=$row['id'];
$response=array();
$response['tags']=$tags;
$response['id']=$id;
}
echo json_encode($response);
?>
的每一次迭代的$回应如果不说明是什么问题,即你希望发生的事情,到底发生了什么,你等等,那么它有什么错误信息我们不可能帮助你。请编辑您的问题并详细说明问题。不,“它不工作”是*不是一个有用的错误描述。我的烤面包机也不工作。 – 2013-03-25 10:27:53
您在while循环中重置$ response。移动'$ response = array();'outside并用'$ response [] = array('tags'=> $ tags,'id'=> $ id)替换内部;'' – Waygood 2013-03-25 10:28:43