Question

我有一个PHP变量$ admin，对应一个复选框。当提交表单并勾选复选框时，将传递该值（设置为1）并且一切正常，但如果未勾选该变量，则会导致未定义的索引错误：

注意：未定义的索引：C：... \ admin.php

中的admin

这是查询db：

的表单的php部分

  if($row == 1)
    {
        echo '<div id="errormsg">This username is already taken</div>';
    }
    else

    {
    $add = mysqli_query($dbcon, "INSERT INTO users (id, firstname, lastname, username, password, admin) VALUES
    (null, '$fname', '$lname', '$user', '$pass', '$admin') ") or die ("Can't insert data");
    echo '<div id="create-success">Successfully added user!</div>';
    }

HTML表单：

  <form action="admin.php" method="post" onSubmit="return validate(this)">
<fieldset>
    <label class="reg">Username *</label> <input type="text" name="user" /><br />
    <label class="reg">Password *</label> <input type="password" name="pass" /><br />
    <label class="reg">Repeat Password *</label> <input type="password" name="rpass" /><br />
    <label class="reg">First name:</label> <input type="text" name="fname" /><br />
    <label class="reg">Last name:</label> <input type="text" name="lname" /><br />
    <label class="reg">Admin?:</label> <input type="checkbox" value="1" name="admin" /><br/>
</fieldset>
    <input type="submit" name="submit" value="Create User" />
</form>

哈有任何想法为什么不选中该框会导致$ admin变量特有的错误？

Answer 1

如果选中复选框，它们实际上只会传递给脚本。

所以你需要检查你的PHP代码是否存在，如下所示：

if ( $_POST && isset( $_POST['admin'] ) {
    $admin = $_POST['admin'];
} else {
    $admin = 0;
}

编辑：

在你的脚本中我会这样做

if($row == 1) {
    echo '<div id="errormsg">This username is already taken</div>';
} else {
    $admin = ( $_POST && isset( $_POST['admin'] ) ? $_POST['admin'] : 0;
    $add = mysqli_query($dbcon, "INSERT INTO users 
               (id, firstname, lastname, username, password, admin) 
         VALUES(null, '$fname', '$lname', '$user', '$pass', '$admin') ") or die ("Can't insert data");
    echo '<div id="create-success">Successfully added user!</div>';
}

Answer 2

首先，您必须检查Post Array中的admin变量。请检查代码。

<?php
if($row == 1)
{
 echo '<div id="errormsg">This username is already taken</div>';
}
else
{
// check is exist in post array
$isAdminCheck = (isset($_POST['admin']) && $_POST['admin'] == 1 )? $_POST['admin']:0;

$add = mysqli_query($dbcon, "INSERT INTO users (id, firstname, lastname, username,   password, admin) VALUES    (null, '$fname', '$lname', '$user', '$pass', '$isAdminCheck') ") or die ("Can't insert data");
echo '<div id="create-success">Successfully added user!</div>';
}
?>

在PHP MYSQL表单中未提交时，Checkbox变量导致错误

2 个答案: