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我已經編寫了此代碼上傳項目,但插入查詢不起作用。Mysql INSERT查詢不起作用,但其他正在工作
<?php
require("connect.php");
if(isset($_POST['submit'])) {
move_uploaded_file($_FILES['file']['tmp_name'],"assets/img/portfolio/".$_FILES['file']['name']);
$filename=$_FILES['file']['name'];
if($filename == ''){
echo "you didn't select any image!";
exit;
}
$title = $_POST['title'];
$desc = $_POST['desc'];
$cat = $_POST['category'];
echo "$title, <br>$filename<br>$desc<br>$cat";
$sql = "INSERT INTO portfolio (Title, Img, Desc, Category) VALUES ('$title' ,'$filename' ,'$desc', '$cat')";
if(mysqli_query($conn, $sql)){
echo "<script>alert('You have Successfully added an item')</script>";
}
else{
echo "<script>alert('Failed. Please try again')</script>";
}
}
mysqli_close($conn);
?>
我已經嘗試在$ sql上分配一個選擇查詢,它的工作原理。請幫助
你能格式化你的PHP代碼嗎?將代碼作爲代碼片段發佈。 –
而不是''alert('失敗,請重試')''使用錯誤報告,http://php.net/manual/en/mysqli.error.php。你也許可以開放SQL注入(這也可能是你當前的問題),使用參數化查詢。 – chris85
我是初學者,我該怎麼做才能避免SQL注入? –