[英]How to check if link (a) submit button is clicked
如何使用PHP檢查是否單擊了此代碼,以便可以在if語句中使用它?
<form id="rating" action="index.php" method="post">
<a href="#" onclick="document.getElementById('rating').submit();">Rating</a>
</form>
因此,如果單擊它,則要使用此查詢:
if () {
$result = mysqli_query($con,"SELECT * FROM users WHERE `approved` = 1 ORDER BY `rating`");
}
您將必須在<input type="hidden" name="hidden_element" value="data"/>
添加輸入,例如<input type="hidden" name="hidden_element" value="data"/>
,否則服務器將沒有POST數據。
然后,在index.php
腳本中,您可以檢查$_POST['hidden_element']
是否已設置。
例如:
if (isset($_POST['hidden_element']) {
$result = mysqli_query($con,"SELECT * FROM users WHERE `approved` = 1 ORDER BY `rating`");
}
因此,由於您當前的表單不提交任何數據,我喜歡表單中是否包含多個按鈕的技術,每個按鈕應具有其屬性名稱和提交類型,並且在php上您檢查if (this button was pressed then) else if (this button was pressed then) else (redirect in none or what ever you need to do when landed)
您的表格,我將ahref更改為帶有按鈕名稱的輸入類型Submit
<form id="rating" action="index.php" method="post">
<input type="submit" name="button"
onclick="document.getElementById('rating').submit();">Rating
</form>
php操作應如下所示,您可以稍后在此處實現ajax調用
if (isset($_POST['button']) === true && empty($_POST['button']) === tre)
{
$result = mysqli_query($con,"SELECT * FROM users WHERE `approved` = 1 ORDER BY `rating`");
}
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