[英]PHP function to check if 2 variables are empty
I have an index page, I want it to include a page called splash.php and not display.php when a user lands on index.php, but once a user does something (sets a variable) ie if a user searches (variable "query") i want it to include display.php and not include splash.php 我有一个索引页,当用户登陆index.php时,我希望它包含一个名为splash.php而不是display.php的页面,但是一旦用户执行了某项操作(设置了变量),即用户搜索了(变量“查询”),我希望它包括display.php而不包括splash.php
What is wrong with this code? 此代码有什么问题?
function hasGet()
{
return !empty($_GET['fact']);
return !empty($_POST['query']);
}
if (hasGet()) {
include("display.php");
}
else {
include("splash.php");
}
This question should be removed 这个问题应该删除
Only the first return statement is executed. 仅执行第一个return语句。 Try:
尝试:
return !empty($_GET['fact']) && !empty($_POST['query']);
A better way to accomplish what you are trying to do is use sessions. 完成会话的一种更好的方法是使用会话。
<?php
session_start();
if (!isset($_SESSION['visited'])) {
$_SESSION['visited'] = true;
include 'splash.php';
} else {
include 'display.php';
}
?>
This way after a user visits index.php
for the first time, $_SESSION['visited']
is set to true and it won't show the splash page throughout their visit. 这样,在用户首次访问
index.php
之后, $_SESSION['visited']
设置为true,并且不会在整个访问过程中显示初始页面。
You cannot have two returns as you are doing. 您不能像做的那样有两个回报。 Try
尝试
return (!empty($_GET['fact']) && !empty($_GET['query']));
You might want to try this... 您可能想尝试一下...
if($_SERVER["SCRIPT_NAME"] == "/index.php"){
include("splash.php");
}else{
include("display.php");
}
2. 2。
if(!empty($_GET["fact"]) || !empty($_POST["query"])){
include("display.php");
}else{
include("splash.php");
}
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