Question

您好，以下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查询来选择WHERE数据等于提交数据的数据。

请参见下文-

import turtle

colors = ["blue", "red", "yellow", "pink"]
NUM_LINES = 100

turtle.hideturtle()
turtle.speed(0)
colors = ["blue", "red", "yellow", "pink"]


for x in range(NUM_LINES):
  color = colors[x//25]
  turtle.left(90)
  turtle.forward(2+(4*x))
  turtle.pencolor(color)

PHP SQL-

include_once'dbh.php';

  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");

        }
    });
}

任何人都可以告诉我这是否是将SELECT查询与AJAX POST数据一起使用的正确方法。

Answer 1

从

更改SQL语句

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

到

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

并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。这样您的整个代码将是

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

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

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

您可以在SQL SELECT查询中使用从AJAX发布的数据吗

1 个答案: