Question

我正在对从PHP函数返回的数组执行JSON解析，但它似乎没有工作。

这是PHP函数：

<!DOCTYPE html>
<html>
<head>
</head>
<body>

<?php

			$bname = $_REQUEST["bname"];

        	$link = mysqli_connect('localhost', 'root', '123'); 

			$servername = "localhost";
			$username = "root";
			$password = "123";
			$dbname = "success";

			// Create connection
			$conn = new mysqli($servername, $username, $password, $dbname);
			// Check connection
			if ($conn->connect_error) {
    			die("Connection failed: " . $conn->connect_error);
			}
			// PHP for execution
			$sql = "SELECT id, bname, bicon, rafrica, rasia, roceania, reurope, rsouthamerica, rnorthamerica, traffic, revenue, profit FROM business LIMIT 1";
			$result = $conn->query($sql);

			if ($result->num_rows > 0) {
    			// output data of each row
    			while($row = $result->fetch_assoc()) {
        			$b3name = $row["bname"]. "<br>";
        			$b3icon = $row["bicon"]. "";
        			$b3rafrica = $row["rafrica"]. "<br>";
        			$b3rasia = $row["rasia"]. "<br>";
        			$b3roceania = $row["roceania"]. "<br>";
        			$b3reurope = $row["reurope"]. "<br>";
        			$b3rsouthamerica = $row["rsouthamerica"]. "<br>";
        			$b3rnorthamerica = $row["rnorthamerica"]. "<br>";
        			$b3traffic = $row["traffic"]. "<br>";
        			$b3revenue = $row["revenue"]. "<br>";
        			$b3profit = $row["profit"]. "<br>";
    			}
			} else {
    			echo "0 results";
			}

			$output = array(
			    'name' => $b3name,
			    'icon' => $b3icon,
			    'traffic' => $b3traffic
			);
			
			echo json_encode($output);

?>
</body>
</html>

这是包含JSON解析的AJAX：

		function loadfacebook1()
		{
			var xmlhttp;
			if (window.XMLHttpRequest)
  			{// code for IE7+, Firefox, Chrome, Opera, Safari
  			xmlhttp=new XMLHttpRequest();
  			}
			else
  			{// code for IE6, IE5
  			xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  			}

			xmlhttp.onreadystatechange=function()
			  {
			  if (xmlhttp.readyState==4 && xmlhttp.status==200)
			    {
			    document.getElementById("b1").innerHTML=xmlhttp.responseText;
			    }
			  } 
 			
 			xmlhttp.open("GET","getfacebook.php",true);
			xmlhttp.send();

			var obj = JSON.parse(xmlhttp.responseText);
			document.getElementById("demo").innerHTML=obj.name + "<br>";
		}

我正在使用

<span id="demo">

显示返回的值，但我需要将obj.name（以及数组的其他一些元素）分配给一个变量，我可以用它来更新页面中的其他内容。任何帮助都会非常感激。

干杯，

威尔

Answer 1

您应该将收到的JSON解析移动到传递AJAX响应时调用的函数（onreadystatechange）

Answer 2

当您从PHP提供JSON结果时，您应该排除HTML代码。尝试删除以下内容，看看是否有帮助：

<!DOCTYPE html>
<html>
<head>
</head>
<body>

和

</body>
</html>

JSON解析没有从数组中获取值

2 个答案: