我正在尝试从 php 发送一个数组(我从 mysql 表中获取到 js)。尽管有很多例子，但我似乎无法使它们中的任何一个发挥作用。到目前为止我已经达到的代码是:
php_side.php
<!DOCTYPE html>
<html>
<body>
<?php
//$q = intval($_GET['q']);
header("Content-type: text/javascript");
$con = mysqli_connect("localhost","root","","Tileiatriki");
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
//mysqli_select_db($con,"users_in_calls");
$sql="SELECT * FROM users_in_calls";
$result = mysqli_query($con,$sql);
/*while($row = mysqli_fetch_array($result)) {
echo $row['User1_number'];
echo "<br/>";
echo $row['User2_number'];
echo "<br/>";
echo $row['canvas_channel'];
echo "<br/>";
}*/
echo json_encode($result);
mysqli_close($con);
?>
</body>
</html>
test_ajax.html
$(document).ready(function(){
$.getJSON('php_side.php', function(data) {
$(data).each(function(key, value) {
// Will alert 1, 2 and 3
alert(value);
});
});
});
这是我使用类似功能的第一个应用程序，所以请耐心等待。
最佳答案
现在您正在发送与 json 响应混合的完整页面标记，这当然是行不通的。
例如，假设您有以下 php 脚本，该脚本假设返回 json 响应:
<div><?php print json_encode(array('domain' => 'example.com')); ?></div>
此页面的响应不会是 json，因为它也会返回包装 div 元素。
您可以将 php 代码移至页面顶部或简单地删除所有 html:
<?php
// uncomment the following two lines to get see any errors
// ini_set('display_errors', 1);
// error_reporting(E_ALL);
// header can not be called after any output has been done
// notice that you also should use 'application/json' in this case
header("Content-type: application/json");
$con = mysqli_connect("localhost","root","","Tileiatriki");
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
$sql="SELECT * FROM users_in_calls";
$result = mysqli_query($con,$sql);
// fetch all rows from the result set
$data = array();
while($row = mysqli_fetch_array($result)) {
$data[] = $row;
}
mysqli_close($con);
echo json_encode($data);
// terminate the script
exit;
?>
关于javascript - 使用ajax和json将数组从php发送到js，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31096743/