当密码和重新键入的密码不匹配时，如何使用php保持用户名 - how to remain the username when password and re-type password did not match using php

how to remain the username when password and re-type password did not match using php

在我的代码中，我在这里有用户名、密码和重新键入密码。我将为我的项目创建一个安全登录。我正在检查密码并重新键入密码，如果不匹配，用户名必须保留在文本框中。

这是我的代码：

<tr>
    <td><label>Username:</label></td>
    <td><input type='text' name='UserName' id='username_input'>
    <div id="feedback" color='red'></div></td>
</tr>
<tr>
    <td><label>Password:</label></td>
    <td><input type="password" name="pasword" required/></td>
</tr>
<tr>
    <td><label>Re-type Password:</label></td>
    <td><input type="password" name="repasword" required/></td>
</tr>
<tr><td>&nbsp;</td></tr>
<tr>
    <td><input type="submit" name="create" value="Submit"></td>
</tr>
<?php
  if(isset($_POST['create'])) {
     $var_user = $_POST['UserName'];
     $var_pass = $_POST['pasword'];
     $var_repass = $_POST['repasword'];
     $user = "user";
     $sqlusr = mysql_query("SELECT * FROM `login` WHERE `username`='$var_user'");
     while($usrrows = mysql_fetch_array($sqlusr)) {
        $dborgid = $orgrows['org_id'];
        $dbusrnme = $usrrows['username'];
     }
    if($var_pass == $var_repass) {
        $infoinsrt = "INSERT INTO `login`(`id`,`username`,`password`, `usertype`)VALUES ('$dborgid','$var_user','$var_pass','$user')";
        $_SESSION['user'] = $var_user;
        $_SESSION['pass'] = $var_pass;
        if (!mysql_query($infoinsrt)) {
            die('Error!' .mysql_error());
        } else {
             echo "<script >('Proceed to next Step.')</script>";
             require 'user_prof.php';    
        }
    } else {
         echo "<script type='text/javascript'>
               alert('Password did Not Match');
               history.back();
               </script>
              ";
    }    
}

<div id="wrapper">
 <form name="form1" name="form1" onsubmit="return validate_login()">
    <tr>
            <td><label>Username:</label></td>
            <td><input type='text' name='UserName' id='username_input'>
            <div id="feedback" color='red'></div></td>
    </tr>
    <tr>
        <td><label>Password:</label></td>
        <td><input type="password" name="pasword" required/></td>
    </tr>
    <tr>
        <td><label>Re-type Password:</label></td>
        <td><input type="password" name="repasword" required/></td>
    </tr>
    <tr><td>&nbsp;</td></tr>
    <tr>
        <td><input type="submit" onclick="validate_login()" name="create" value="Submit"></td>
    </tr>
    <span style="color:red;font-size:14px;" id='error'> </span> <br>
 </form>
</div>
<script src="//code.jquery.com/jquery-1.11.2.min.js"></script>
<script>
  function validate_login() {
    var name = document.forms["form1"]["UserName"].value;
    var password = document.forms["form1"]["pasword"].value;
    var password1 = document.forms["form1"]["repasword"].value;
    if(password1 != password) {
         document.getElementById("error").innerHTML = "Password's does not match!!" 
         return false;
    } else {
         return true;
    }
}
</script>
<noscript>
   <style type="text/css">
        #wrapper {display:none;}
   </style>
     Your browser JavaScript is disabled!..Please enable javascript to see the  content..
</noscript>

您可以使用javascript，而不是使用php进行检查，它使过程更容易。希望这能帮助你。。。

您可以echo它。

<tr>
    <td><label>Username:</label></td>
    <td><input type='text' name='UserName' id='username_input' value="<?php echo $var_user;?>">
    <div id="feedback" color='red'></div></td>
</tr>

如果未设置，则使$var_user分配给$var_user = ''。