Question

我是jquery新手，使用jquery将数据插入mysql数据库时遇到一些麻烦。这是我的html文件

<!DOCTYPE html>
<html>
<head>
<title>
staff registration
</title>
<body>
<form>
Staff Id:<input type="text" id="staffid"><br>
Password:<input type="password" id="password1"><br>
Re-enter Password:<input type="password" id="password2"><br>
Email:<input type="email"   id="email"><br>
Gender:<input type="text" id="gender"><br>
Qualification:<input type="text" id="qualification"><br>
Course 1:<input type="text" id="course1"><br>
Course 2:<input type="text" id="course2"><br>
Course 3:<input type="text" id="course3"><br><br>
<input type="submit" id="submit" value="CREATE">
</form>
<script src="jquery.min.js"></script>
<script>
$("#submit").click(function(){
 var staffid=$("#staffid").val();
 var password1=$("#password1").val();
 var password2=$("#password2").val();
 var email=$("#email").val();
 var gender=$("#gender").val();
 var qualification=$("#qualification").val();
 var course1=$("#course1").val();
 var course2=$("#course2").val();
 var course3=$("#course3").val();
 $.post("insert.php",{si:staffid,pwd1:password1,pwd2:password2,
 email:eml,gender:gen,qualification:qal,course1:c1,course2:c2,course3:c3},
 function(data){
     alert(data);
 }); 
});
</script>
</body>
</html>

这是我的名为insert.php的php文件

<?php
 $con=new mysqli("localhost","root","","flash");
    $si=$_POST['si'];
    $pwd1=$_POST['pwd1'];
    $pwd2=$_POST['pwd2'];
    $eml=$_POST['eml'];
    $gen=$_POST['gen'];
    $qal=$_POST['qal'];
    $c1=$_POST['c1'];
    $c2=$_POST['c2'];
    $c3=$_POST['c3'];
    $sql="INSERT INTO staff(staff_id,password,email_id,gender,qualification,course_1,course_2,course_3) VALUES('$si','$pwd2','$eml','$gen','$qal','$c1','$c2','$c3')";
    $con->query($sql);

?>

当我单击“提交”按钮时什么都不会发生

请任何人帮助我解决此问题

Answer 1

尝试使用非精简版（可在http://jquery.com/download/上找到），例如：

https://code.jquery.com/jquery-3.3.1.min.js

在制作帖子的json时您也遇到了问题，这是更正后的版本

<!DOCTYPE html>
<html>
<head>
<title>
staff registration
</title>
<body>
<form>
Staff Id:<input type="text" id="staffid"><br>
Password:<input type="password" id="password1"><br>
Re-enter Password:<input type="password" id="password2"><br>
Email:<input type="email"   id="email"><br>
Gender:<input type="text" id="gender"><br>
Qualification:<input type="text" id="qualification"><br>
Course 1:<input type="text" id="course1"><br>
Course 2:<input type="text" id="course2"><br>
Course 3:<input type="text" id="course3"><br><br>
<input type="submit" id="submit" value="CREATE">
</form>
<script
  src="https://code.jquery.com/jquery-3.3.1.min.js"></script>
<script>
$("#submit").click(function(){
 var staffid=$("#staffid").val();
 var password1=$("#password1").val();
 var password2=$("#password2").val();
 var email=$("#email").val();
 var gender=$("#gender").val();
 var qualification=$("#qualification").val();
 var course1=$("#course1").val();
 var course2=$("#course2").val();
 var course3=$("#course3").val();
 $.post("insert.php",{si:staffid,pwd1:password1,pwd2:password2,
 eml:email,gen:gender,qal:qualification,c1:course1,c2:course2,c3:course3},
 function(data){
     alert(data);
 }); 
});</script>
</body>
</html>

使用jQuery将数据插入MySQL时出错

1 个答案: