Question

我在我的网站空间上运行一个用于Android应用程序的脚本。虽然，它似乎没有用。在某些网站空间中，当我尝试访问它时，我得到一个空白站点，而在其他网站空间中，我收到500个内部服务器错误。虽然这个脚本应该正常工作，因为它曾经这样做。

它应该从表中获取所有信息并在一些INNER JOIN之后回显它。即使在一些mysqli_error检查和一些回声检查变量等之后，我也无法跟踪错误。

变量$ con确实存在，连接应该正常工作。我刚把它删除了堆栈溢出^^。

 <?php


$subId = 1;


$sql = "SELECT articles.a_id, sciences.science, articles.title, articles.content, login.username, articles.date, articles.viewed, articles.timestamp FROM articles 
                INNER JOIN sciences ON articles.s_id = sciences.s_id 
                INNER JOIN login ON articles.author = login.id
                WHERE articles.s_id = ".$subId."
                ORDER BY timestamp DESC";

if (!$res = mysqli_query($con,$sql)) {
    echo "FAIL";
    echo mysqli_error($con);
}
else {

$result = array();

while($row = mysqli_fetch_array($res)){
array_push($result,
array('a_id'=>$row[0],
'science'=>$row[1],
'title'=>$row[2],
'content'=>$row[3],
'author'=>$row[4],
'date'=>$row[5],
'viewed'=>$row[6],
'timestamp'=>$row[7]
));
}

echo json_encode(array("result"=>$result));
}
mysqli_close($con);


?>

编辑：我更新了我的代码：

    <?php


$subId = 1;

$con = mysqli_connect("HOST", "USER", "PASSWORD", "DATABASE");

if (!$con)
{
echo "Cant't connect to MySQL.<br>";
echo "Debug: " . mysqli_connect_errno()
}


$sql = "SELECT articles.a_id, sciences.science, articles.title, articles.content, login.username, articles.date, articles.viewed, articles.timestamp FROM articles 
INNER JOIN sciences ON articles.s_id = sciences.s_id 
INNER JOIN login ON articles.author = login.id
WHERE articles.s_id = ".$subId."
ORDER BY timestamp DESC";

if (!$res = $con->query($sql))
{
    echo "FAIL";
}
else
{

    $result = array();

    while($row = $con->fetchArray($res))
    {
        array_push($result); //use the query to rename fields, if needed
    }

    echo json_encode(array("result"=>$result));
}



?>

Answer 1

我认为问题的一部分是你混淆了mysql和mysqli函数。你在程序上使用了mysqli，即：mysqli_fetch_array（），它并不存在。你必须转换为mysql，或者使用像这样的mysqli函数：

$subId = 1;

$con = mysqli_connect('host', 'user', 'pass', 'dbname');

if (!$con)
{
echo "Cant't connect to MySQL.<br>";
echo "Debug: " . mysqli_connect_errno()
}


$sql = "SELECT articles.a_id, sciences.science, articles.title, articles.content, login.username, articles.date, articles.viewed, articles.timestamp FROM articles 
INNER JOIN sciences ON articles.s_id = sciences.s_id 
INNER JOIN login ON articles.author = login.id
WHERE articles.s_id = ".$subId."
ORDER BY timestamp DESC";

if (!$res = $con->query($sql))
{
    echo "FAIL";
}
else
{

    $result = array();

    while($row = $con->fetchArray($res))
    {
        array_push($result); //use the query to rename fields, if needed
    }

    echo json_encode(array("result"=>$result));
}

Answer 2

将时间戳更改为$logged = auth()->attempt( ['email' => $request['email'], 'password' => $request['password'], 'role' => 'user']); if($logged) { return $next($request); // probably it won't work. This is fine in middleware but not in controller } else{ return redirect()->back(); }，因为它是保留字添加反引号

Answer 3

尝试在PHP中使用错误报告。通过在代码开头添加以下行来启用错误报告。

ini_set('display_startup_errors', 1);
ini_set('display_errors', 1);
error_reporting(-1);

Answer 4

显然你忘记了

$con = mysqli_connect("localhost","my_user","my_password","my_db");

在脚本开头。

PHP脚本似乎不起作用

4 个答案: