jquery - 从 URL 获取变量

Question

我正在检查是否超出了任务必须完成的时间，以发送警报。Jquery 代码:

function session_checking1()
{
    $.post( "./alertaposicionamento", function( data ) {
        if(data == "-1")
        {
           alert('Tem posicionamentos em atraso!');
        }
    });
}
var validateSession1 = setInterval(session_checking1, 10000);

在 alertaposicionamento 页面上，我有以下 php:

$result1 = mysqli_query($conn, $query2);  
while($row = mysqli_fetch_array($result1))  
      { 
        $teste = $row["codigo"];
        $teste1 = $row["Colaborador"];
        $teste2 = $row["FimTarefa"];
        $teste3 = $row["TipoPeriodicidade"];
        $teste4 = $row["Tempo"];
        $teste5 = $row["Ala"];
        $teste8 = $row["nome"];
      }

$query = "SELECT iduser, hostname FROM raddb.sessoes
WHERE datafim IS NULL AND hostname = '$teste5'";
$result = mysqli_query($conn, $query);  
while($row1 = mysqli_fetch_array($result))  
      { 
        $teste6 = $row1["iduser"];
        $teste7 = $row1["hostname"];
      }         

if($teste4 > $teste3 AND $teste5 == $teste7 AND $teste6 == $_SESSION['usuarioId'])
{
    //expired
    echo "-1";
}
else
{
    //not expired
    echo "1";
}

它工作正常并且符合我的意图。我想要进行改进，我想要获取 $test8 变量的值，通过 URL 发送它，并将该变量的值添加到我在 Jquery< 内的消息中 alert('Tem posicionamento em atraso do (以及 $test8 变量的值)!');

Answer 1

您需要将值与您的响应一起发回

header('Content-Type: application/json');
if($teste4 > $teste3 AND $teste5 == $teste7 AND $teste6 == $_SESSION['usuarioId'])
{
    //expired
    echo json_encode(["status"=>"-1","value"=>$teste8]);
}
else
{
    //not expired
     echo json_encode(["status"=>"1","value"=>$teste8]);
}

然后将你的js更改为

if(data.status == "-1")
        {
           alert('Tem posicionamento em atraso do '+data.value);
        }