嗨,下面的ajax将数据提交到URL。我已经在PHP文件中检索了数据,并且想编写一个SQL查询以选择WHERE数据等于已提交数据的数据。

见下文 -

  function loadJobRequests() {


      //AJAX code to submit form.
     $.ajax({
        type: "POST",
        url: "http://localhost:8888/EduSubOct/json-data-
        jobrequests.php",
        data: { userEmail: localStorage.getItem("email")} ,
        cache: false,
        success: function(html) {
            alert("Information Entered Successfully");

        }
    });
}


PHP SQL-

include_once'dbh.php';

//Here we fetch the data from the URL that was passed from our HTML
   form
   $userEmail = $_POST['userEmail'];

   $sql =  "SELECT * FROM jobRequest WHERE email='$userEmail'";

    $result = mysqli_query($conn, $sql);


谁能告诉我这是否是对AJAX POST数据使用SELECT查询的正确方法。

最佳答案

从更改SQL语句

$sql =  "SELECT * FROM jobRequest WHERE email='$userEmail'";




$sql =  "SELECT * FROM jobRequest WHERE email='".$userEmail."'";


并且始终尝试检查该值是否已设置,有时没有发送任何值,并且您的php脚本将生成错误。所以你的整个代码是

if (isset( $_POST['userEmail'])) {
    $userEmail = $_POST['userEmail'];

    $sql =  "SELECT * FROM jobRequest WHERE email='".$userEmail."'";

    $result = mysqli_query($conn, $sql);
}

10-07 12:26
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