我用PHP之前,從來沒有遇到這個問題....IF ELSE在PHP中不執行,調試,仍然無法找到問題
我有過自己的價值單選按鈕。提交表單時,我已驗證傳遞的值是否與我的代碼中的if語句條件匹配。但是,它不會執行該路徑。
例如,選擇城市單選按鈕時,不會執行與城市相關的if條件。我通過在if語句之前放置一個echo語句進行調試,並且它匹配if語句中的值,因此我對它爲什麼不執行的混淆。
如果有人能幫助我會很感激......我很困惑。
繼承人的HTML:
<html>
<head>
<title>Lab 5</title>
<link rel='stylesheet' href='SearchForm.css' type='text/css' media='all' />
</head>
<body>
<p>You may use % as a wildcard character in your search.</p>
<form method='POST' action='index.php?type=search'>
<input type='text' name='search_param' value='' />
<input type='radio' name='search-type' value='City' id='city' />
<label for='city'>City</label>
<input type='radio' name='search-type' value='Country' id='country' />
<label for='country'>Country</label>
<input type='radio' name='search-type' value='Language' id='language' />
<label for='language'>Language</label>
<br /><br />
<input type='submit' name='submit' value='Search' />
</form>
<p>Or, <a href='index.php?type=insert'>perform an insertion.</a></p>
</body>
</html>
和PHP:
<?php
include 'connect.php';
$table = $_POST['search-type'];
$search = $_POST['search_param'];
if($table != NULL)
{
echo '<table>';
if($table == 'City')
{
$query = "SELECT * FROM city WHERE name ILIKE $1 ORDER BY name;";
$stmt = pg_prepare($connection, "city", $query);
$result = pg_execute($connection, "city", array($search));
while($row = pg_fetch_assoc($result))
{
cityTable($row);
}
echo '</table>';
}
}
else
{
echo 'Please select a table to query!';
}
正如你所看到的,我甚至從我如果狀態拆下變量$search
,而不是直接去後。
當我回應$search
它產生「城市」,所以.....請幫助!
試圖改變名稱='搜索型'with name ='search-type []' – JackTurky
''city'!='City''? –