在php中解析json值并显示在表中
这是我的html在php中解析json值并显示在表中,php,html,json,Php,Html,Json,这是我的html <html> <head> <script src="js/jquery.js"></script> </head> <body> <input type="text" name="query" id="queryBox"> <button id="load_basic" value = "button">search</button><br/> &l
<html>
<head>
<script src="js/jquery.js"></script>
</head>
<body>
<input type="text" name="query" id="queryBox">
<button id="load_basic" value = "button">search</button><br/>
<div id="result"></div>
</body>
<script>
$.ajaxSetup ({
cache: false
});
var ajax_load = "<img src='images/ajax-loader.gif' alt='loading...' />";
$("#load_basic").click(function(){
var query = $("#queryBox").val();
var loadUrl = "http://localhost:8983/solr/collection1/select?q=" + query +"&fl=title&wt=php&indent=true";
$("#result").html(ajax_load).load(loadUrl);
});
</script>
</html>
如何解析此json并仅获取标题中的值
<?php
$query = "http://localhost:8983/solr/collection1/select?q=" . $_GET['q'] ."&fl=title&wt=json&indent=true";
$response = file_get_contents($query);
echo $response;
?>
在JS中,要解析div中的JSON,可以这样做
var myJSON=jQuery.parseJSON(jQuery('#result').text());
var title=myJSON.response.docs[0].title;
console.log(title);
var myJSON=jQuery.parseJSON(jQuery('#result').text());
var title=myJSON.response.docs[0].title;
console.log(title);