PHP返回whyever HTML标记
好的,我有以下php代码:PHP返回whyever HTML标记,php,html,xmlhttprequest,Php,Html,Xmlhttprequest,好的,我有以下php代码: <!doctype html> <html> <body> <?php $q = intval($_GET['q']); $con = mysqli_connect('localhost', 'root', '', 'testDB'); if(!$con){ die('Could not connect: '. mysqli_error($con));
<!doctype html>
<html>
<body>
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost', 'root', '', 'testDB');
if(!$con){
die('Could not connect: '. mysqli_error($con));
}
mysqli_select_db($con, "testDB");
$query = "SELECT * FROM `aTable`;";
$result = mysqli_query($con,$query);
$row = mysqli_fetch_array($result);
echo $row[$q];
mysqli_close($con);
exit();
?>
</body>
</html>
不管怎样,控制台会记录
<!doctype html>
<html>
<body>
-3, -1, -2, 0, 2, 1, 2, 4, 5, 3, 4, 2, 4, 5, 6, 2, 3, 1, 3, 4, 1, -1, -3, -1
-3, -1, -2, 0, 2, 1, 2, 4, 5, 3, 4, 2, 4, 5, 6, 2, 3, 1, 3, 4, 1, -1, -3, -1
我不知道为什么,但它返回这些HTML标记,我不想要(只有从-3到-1的值[带逗号])
如何删除(未关闭的)HTML标记
谢谢你的回答 通过ajax调用
script.php
时,响应将包含文件中的所有html标记,请确保将它们全部删除。此外,您可能希望使用函数将php对象作为字符串发送到客户端,下面是一个示例:
script.php
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost', 'root', '', 'testDB');
if(!$con){
die('Could not connect: '. mysqli_error($con));
}
mysqli_select_db($con, "testDB");
$query = "SELECT * FROM `aTable`;";
$result = mysqli_query($con,$query);
$row = mysqli_fetch_array($result);
echo json_encode($row[$q], true);
mysqli_close($con);
exit();
从文件中删除HTML标记。当然可以。在文本编辑器中打开PHP文件。移除你不想要的东西。保存文件。(注意:标记是“未关闭的”,因为您调用了exit()
,它在关闭标记之前终止脚本。)真的这样问吗?
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost', 'root', '', 'testDB');
if(!$con){
die('Could not connect: '. mysqli_error($con));
}
mysqli_select_db($con, "testDB");
$query = "SELECT * FROM `aTable`;";
$result = mysqli_query($con,$query);
$row = mysqli_fetch_array($result);
echo json_encode($row[$q], true);
mysqli_close($con);
exit();