[英]How can I use localStorage in php with AJAX to update my database?
我需要在PHP文件中使用localStorage值來更新數據庫中的值。 我知道我需要ajax來實現這一點,但我無法使其正常工作。
我的localStorage項命名為option,並且存在(在瀏覽器中檢出並將值存儲在div中 )
$(document).ready(function(){
$.ajax({
type: "POST",
url: "activity.php",
data: { storageValue: localStorage.getItem('option') },
success: function(data){
alert('success');
}
});
});
PHP示例:
$option = $_POST['storageValue'];
mysql_query("...SET x = '".$option."'...");
echo 'Update complete';
我沒有看到任何帖子數據,也沒有得到回復。
謝謝!
您的頁面:
<form>
<input type="hidden" value="thedatayouwanttopost" id="option"/>
</form>
<script src="Your_Js_File.js"></script>
您的JS文件:
document.getElementById('option').submit(); // This submits the form
var storageValue = $('#option').val(); // This gets the value of the form once it has been posted to the .php file
$(document).ready(function(){
$.ajax({
type: "POST",
url: "activity.php",
data: { storageValue:storageValue },
success: function(data){
alert('success');
}
});
return false; // This stops the page from refreshing
});
您的PHP文件將數據回調到AJAX並顯示警報(activity.php):
...
$option = $_POST['storageValue']; // This is the post value of your hidden input on your page
mysql_query("...SET x = '".$option."'...");
?><input type="hidden" id="option" value="<?php echo $option; ?>"><?
// The input above posts the value of 'option' on your page back to your Ajax and spits out the request.
...
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