PHP-更新脚本不工作
我的更新脚本有问题,我的脚本有问题吗?因为每当我运行它时,它只显示echo“ERROR”。这是我的更新脚本:PHP-更新脚本不工作,php,Php,我的更新脚本有问题,我的脚本有问题吗?因为每当我运行它时,它只显示echo“ERROR”。这是我的更新脚本: <?PHP $con = mysqli_connect("localhost","root","","testuser"); $choosed2 = $_POST['choosed2']; $creditnew = $_POST['creditnew']; $debitnew = $_POST['debitnew']; $newid =
<?PHP
$con = mysqli_connect("localhost","root","","testuser");
$choosed2 = $_POST['choosed2'];
$creditnew = $_POST['creditnew'];
$debitnew = $_POST['debitnew'];
$newid = $choosed2 - 1;
$bal = "SELECT balance FROM `bal` WHERE id=$newid;";
$chek = mysqli_query($con,$bal);
$row = mysqli_fetch_assoc($chek);
$old_balance = $row['balance'];
if ($old_balance == NULL){
$old_balance = 0;
}
$balancenew = $old_balance - $creditnew + $debitnew;
$up = "UPDATE `bal` SET debit='$debitnew' credit='$creditnew' balance='$balancenew' WHERE id='$choosed2'";
if(mysqli_query($con,$up))
{
echo "<div class='form' style='text-align:center; float:center;'><div style='font-weight:bold; font-size:20px; margin:10px'>UPDATE SUCCESS</div><br/><a href='index.php' style='margin:5px;' class='tombol'><< BACK</a><a href='Input.php' style='margin:5px;' class='tombol'>INPUT AGAIN</a></div>";
}else{
echo "<div class='form'><h3>ERROR</h3><br/><a href='balance.php'>Input again</a>";
}
?>
您刚才在SQL语句中遗漏了逗号(,
)。您的SQL查询将被删除
$up = "UPDATE `bal` SET debit='$debitnew', credit='$creditnew', balance='$balancenew' WHERE id='$choosed2'";
值之间缺少逗号。此外,ERROR
是一种知道错误的可怕方式。用于获取正确的错误消息<代码>设置借方='$debitnew',贷方='$creditnew'
etc1。添加错误报告(E\u ALL);ini设置(“显示错误”,1)代码>刚启动使用link:-的代码。还要尝试阅读准备好的语句,并在代码中使用它,以防止代码SQL注入