PHP 的致命错误
Fatal error of PHP
我正在做一个管理面板,我正在做一个用于添加用户的表单。我的代码如下:
<form class="form-horizontal">
<fieldset>
<!-- Form Name -->
<legend>Add user</legend>
<!-- Prepended text-->
<div class="form-group">
<label class="col-md-4 control-label" for="newuser">Username</label>
<div class="col-md-4">
<div class="input-group">
<span class="input-group-addon">@</span>
<input id="newuser" name="newuser" class="form-control" placeholder="Username" type="text" required="">
</div>
<p class="help-block">Put the new users username here</p>
</div>
</div>
<!-- Password input-->
<div class="form-group">
<label class="col-md-4 control-label" for="userpw">New user password</label>
<div class="col-md-4">
<input id="userpw" name="userpw" type="password" placeholder="Password" class="form-control input-md" required="">
<span class="help-block">Enter new user's password here!</span>
</div>
</div>
<!-- Select Basic -->
<div class="form-group">
<label class="col-md-4 control-label" for="permissions">Choose permissions</label>
<div class="col-md-4">
<select id="permissions" name="permissions" class="form-control">
<option value="1">System administrator</option>
<option value="2">Editor</option>
<option value="3">Developer</option>
<option value="4">Normal user</option>
</select>
</div>
</div>
<!-- Button -->
<div class="form-group">
<label class="col-md-4 control-label" for="btn_send">Send request to server</label>
<div class="col-md-4">
<button id="btn_send" name="btn_send" class="btn btn-primary">Send it!</button>
</div>
</div>
</fieldset>
</form>
<?php
$newuser = $_GET['newuser'];
$userpw = $_GET['userpw'];
$permission = $_GET['permissions'];
if(isset($newuser and $userpw and $permission) {
$sql = "INSERT INTO users(username,userpw,permission)VALUES('$newuser', '$userpw','$permission')";
mysql_query($sql);
}
else {
echo "error";}
然后我得到:
致命错误:C:\xampp\htdocs\lumino\usermanagement.php 第 95[=12 行中的表达式结果无法使用 isset()(您可以使用 "null !== expression" 代替) =]
无论如何我都无法修复它,玩了大约 5 个小时。
它正在执行 $_GET 方法,所以我得到 url 就像:
mysite.com/usermanagement.php?newuser=Berkay&userpw=18042003&permissions=3&btn_send=
使用 isset
,您不能在其中包含多个项目:
if(isset($newuser and $userpw and $permission) {
需要成为:
if(isset($newuser) && isset($userpw) && isset($permission)) {
或者:
if(isset($newuser, $userpw, $permission)) {
你只是错过了一个 )
和一组 ,
:
if(isset($newuser, $userpw, $permission)) {
$sql = "INSERT INTO users(username,userpw,permission)VALUES('$newuser', '$userpw','$permission')";
mysql_query($sql);
}
我正在做一个管理面板,我正在做一个用于添加用户的表单。我的代码如下:
<form class="form-horizontal">
<fieldset>
<!-- Form Name -->
<legend>Add user</legend>
<!-- Prepended text-->
<div class="form-group">
<label class="col-md-4 control-label" for="newuser">Username</label>
<div class="col-md-4">
<div class="input-group">
<span class="input-group-addon">@</span>
<input id="newuser" name="newuser" class="form-control" placeholder="Username" type="text" required="">
</div>
<p class="help-block">Put the new users username here</p>
</div>
</div>
<!-- Password input-->
<div class="form-group">
<label class="col-md-4 control-label" for="userpw">New user password</label>
<div class="col-md-4">
<input id="userpw" name="userpw" type="password" placeholder="Password" class="form-control input-md" required="">
<span class="help-block">Enter new user's password here!</span>
</div>
</div>
<!-- Select Basic -->
<div class="form-group">
<label class="col-md-4 control-label" for="permissions">Choose permissions</label>
<div class="col-md-4">
<select id="permissions" name="permissions" class="form-control">
<option value="1">System administrator</option>
<option value="2">Editor</option>
<option value="3">Developer</option>
<option value="4">Normal user</option>
</select>
</div>
</div>
<!-- Button -->
<div class="form-group">
<label class="col-md-4 control-label" for="btn_send">Send request to server</label>
<div class="col-md-4">
<button id="btn_send" name="btn_send" class="btn btn-primary">Send it!</button>
</div>
</div>
</fieldset>
</form>
<?php
$newuser = $_GET['newuser'];
$userpw = $_GET['userpw'];
$permission = $_GET['permissions'];
if(isset($newuser and $userpw and $permission) {
$sql = "INSERT INTO users(username,userpw,permission)VALUES('$newuser', '$userpw','$permission')";
mysql_query($sql);
}
else {
echo "error";}
然后我得到:
致命错误:C:\xampp\htdocs\lumino\usermanagement.php 第 95[=12 行中的表达式结果无法使用 isset()(您可以使用 "null !== expression" 代替) =]
无论如何我都无法修复它,玩了大约 5 个小时。 它正在执行 $_GET 方法,所以我得到 url 就像:
mysite.com/usermanagement.php?newuser=Berkay&userpw=18042003&permissions=3&btn_send=
使用 isset
,您不能在其中包含多个项目:
if(isset($newuser and $userpw and $permission) {
需要成为:
if(isset($newuser) && isset($userpw) && isset($permission)) {
或者:
if(isset($newuser, $userpw, $permission)) {
你只是错过了一个 )
和一组 ,
:
if(isset($newuser, $userpw, $permission)) {
$sql = "INSERT INTO users(username,userpw,permission)VALUES('$newuser', '$userpw','$permission')";
mysql_query($sql);
}