我正在使用PHP文件中的sql請求,但我遇到了問題。我GET fonction作品:SQL更新函數
if(isset($_POST['week']) && !empty($_POST['week']) && isset($_POST['location']) && !empty($_POST['location']))
{
$week = $_POST['week'];
$location = $_POST['location'];
$sql = 'SELECT *
FROM myDataBaseName
WHERE myDataBaseName_week='.$week.'
AND myDataBaseName_location="'.$location.'"';
}
我的更新功能不看工作:
if (isset($_POST['week']) && !empty($_POST['week']) && isset($_POST['location']) && !empty($_POST['location']) && isset($_POST['number']) && !empty($_POST['number']))
{
$week = $_POST['week'];
$location = $_POST['location'];
$numberForUpdate = $_POST['number'];
$sql = 'UPDATE *
FROM myDataBaseName
SET myDataBaseName_numbers ='.$numberForUpdate.'
WHERE myDataBaseName_week='.$week.'
AND myDataBaseName_location="'.$location.'"';
}
誰能幫我得到一個正確的更新功能嗎?
謝謝!
你應該考慮在連接前使用mysqli_real_escape_string'$周= $ _ POST [ '周'];'入'$周= mysqli_real_escape_string($ _ POST [ '周']);' – 2014-09-20 13:36:53
'UPDATE myDataBaseName SET myDataBaseName_numbers ='。 $ numberForUpdate「。 WHERE myDataBaseName_week ='。$ week。' AND myDataBaseName_location =「'。$ location。'」'; – 2014-09-20 13:36:59
語法爲'UPDATE SET SET ...',而不是'UPDATE * FROM SET SET ...' - 在打開'<?php'標記後立即向文件頂部添加錯誤報告 'error_reporting (E_ALL); ini_set('display_errors',1);'看看它是否產生任何東西。 – 2014-09-20 14:21:07