[英]PHP AND IF statement
即時通訊從一個簡單的HTML頁面調用php頁面,我找不到語法錯誤,這是html
<html>
<body style="background-color:#990000;">
<h5 align="center" style="font-family:tahoma;color:white;font-size:50px;"> Welcome! </h5>
<form align="center" method="post" action="php_site.php">
First name: <input type="text" name="firstname"><br>
Last name: <input type="text" name="lastname"><br>
<input type="submit" value="Search!">
</form>
</body>
</html>
這是PHP
<?php
if (isset($_POST['firstname']) && (isset($_POST['lastname'])))
{
echo "Welcome $_POST['firstname']";
echo "This is your last name $_POST['lastname']";
}
else
{
echo "This is Empty!";
}
?>
謝謝!
字符串中對$ _POST的嵌入式引用的語法不太正確。 解決它的一些方法是:
將它們用{}
括起來:
echo "Welcome {$_POST['firstname']}"; echo "This is your last name {$_POST['lastname']}";
或者,刪除單引號:
echo "Welcome $_POST[firstname]"; echo "This is your last name $_POST[lastname]";
或者,將字符串連接與一起使用.
而不是嵌入:
echo "Welcome " . $_POST['firstname']; echo "This is your last name " . $_POST['lastname'];
和/或,首先將值拉入純名稱的變量中:
$firstname = $_POST['firstname']; $lastname = $_POST['lastname']; echo "Welcome $firstname"; echo "This is your last name $lastname";
請參閱有關字符串中的變量解析的文檔。
嘗試這個:
<?php
if (isset($_POST['firstname']) && isset($_POST['lastname']))
{
echo "Welcome ".$_POST['firstname'];
echo "This is your last name ".$_POST['lastname'];
}
else
{
echo "This is Empty!";
}
?>
<?php
if (isset($_POST['firstname']) && isset($_POST['lastname']) )
{
echo "Welcome ".$_POST['firstname'];
echo "This is your last name $_POST['lastname']";
}
else
{
echo "This is Empty!";
}
?>
首先,您應該將$_POST
值設置為變量。
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
然后,確保使用串聯標點符號從變量中拆分字符串.
:
echo "Welcome" . $firstname;
而不是echo "Welcome $_POST['firstname']";
另外,請發布您遇到的語法錯誤。 如果PHP崩潰了(白屏死機),請添加ini_set("display_errors", "1");
為了將錯誤寫入屏幕,然后將錯誤輸出。
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