使用插入查询进入函数时表单不提交?

时间:2016-03-29 12:15:34

标签: php jquery ajax forms

在函数外部使用insert查询时它的工作原理。但是当使用功能时,它不起作用。

php代码:

 <?php

function registerusers(){
  $username =$_POST['username'];
  $password =$_POST['password'];
  $error = "Your Login Name or Password is invalid";
  echo "$username";
  $sql = "SELECT id FROM users WHERE username = '$username' and password = '$password'";
  $result = mysqli_query($conn,$sql);
  $row = mysqli_fetch_array($result,MYSQLI_ASSOC);
  $count = mysqli_num_rows($result);
  if($count == 1) {
  ?>
  <script type="text/javascript">
  $("div#ack").html("Successfully");
  </script>
  <?php
  }else {
  ?>
  <script type="text/javascript">
  $("div#ack").html("Invalid user and password");
  </script>
  <?php
  }
}
registerusers()

jquery代码:

$("button#submit").click( function() {
    if( $("#username").val() == "" || $("#password").val() == "" ){
    $("div#ack").html("Please enter both username and password");
    }else{
    $.post( $("#myForm").attr("action"),
    $("#myForm :input").serializeArray(),
    function(data) {
    $("div#ack").html(data);
    });
    $("#myForm").submit( function() {
    return false;   
    });
    }
});
<form id="myForm" action="functions.php" method="POST">
    username: <input type="text" name="username" /><br />
    password: <input type="password" name="password" /><br />
    <button id="submit">Login</button>
</form>

我同时使用方法按钮类型按钮和按钮类型提交

0 个答案:

没有答案