我正在对PHP函数返回的数组执行JSON解析，但它似乎不起作用。
以下是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>以下是包含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">以显示返回的值，但我需要将obj.name（以及数组的一些其他元素）分配给一个变量，我可以使用该变量来更新页面中的其他内容。如有任何帮助，我们将不胜感激。
干杯，
是否
您应该将接收到的JSON的解析转移到AJAX响应（onreadystatechange）传递时调用的函数
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，看看它是否有帮助：
<!DOCTYPE html>
<html>
<head>
</head>
<body>
和
</body>
</html>