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php - jQuery ajax (JSON) 从 PHP 获取数组工作正常,但在网络服务器上不行

转载 作者:行者123 更新时间:2023-11-29 13:56:36 24 4
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我发现了一些类似的问题,但没有解决方案解决这个问题。

function loadProject(id) {
$.ajax({
url: 'loadDrumsetData.php',
type: 'GET',
data: {
i: id
},
dataType: 'JSON',
success: function (e) {
pushLoadedData(e.bank); //create the JavaScript array
},
error: function (request, textStatus, errorThrown) {
console.log(request, textStatus, errorThrown);
}
});
}

我尝试从 loadDrumsetData.php 获取数组结果。在我本地的 apache 上它工作正常。但是在我的网络服务器(apache)上我收到这个解析器错误:

SyntaxError: JSON.parse: unexpected end of data

loadDrumsetData.php:

<?php
header('Content-type: application/json; charset=UTF-8');
error_reporting(-1);

$i=$_GET["i"];

$con=mysqli_connect("localhost","userxxx","xxxxx","drumpcdata");

if (mysqli_connect_errno())

echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con," SELECT * FROM `session_".$i."` ORDER BY `pID` ASC ");

$soundArray = array();
$bankArray = array();

while($row = mysqli_fetch_array($result))
{
$pid = $row['pID'];
$r = $row['Row'];
$sound = $row['Sound'];
$number = $row['Number'];
$x = $row['X'];
$y = $row['Y'];
$w = $row['W'];
$h = $row['H'];
$spr = $row['Sprite'];
$pressed = $row['Pressed'];
$bankArray = [];
array_push($soundArray, [$pid, $r, $sound ,$number,$x,$y,$w,$h,$spr, $pressed]);
array_push($bankArray, $soundArray);
}

mysqli_close($con);

$encoded = json_encode(array("bank" => $bankArray));
echo $encoded;
?>

如果您对此主题有任何帮助,我将不胜感激。知道问题是什么吗?

最佳答案

您确定正确连接到mysql数据库吗?

首先,如果连接失败,不会停止处理:

if (mysqli_connect_errno())
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

并尝试进行查询等...

由于失败,您的服务器发送字符串而不是 json,它会炸毁一切。

if (mysqli_connect_errno()){
echo json_encode(
array(
"Success" => false,
"Reason" => "Failed to connect to MySQL: ".mysqli_connect_error()
)
);
exit(0);
}

请尝试向您的脚本添加一些故障控制。

查询也可能失败。

关于php - jQuery ajax (JSON) 从 PHP 获取数组工作正常,但在网络服务器上不行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15761999/

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