我无法弄清楚我的SQL查询有什么问题。
我有一个相当标准的形式，只要我用lorem ipsum填充字段就可以正常工作。也适用于空字段。当我输入信息时，我想提交它失败。以下是我杀死脚本时mysql错误告诉我的内容：
＆＃39;您的SQL语法有错误;查看与您的MySQL服务器版本相对应的手册，以获得正确的语法，以便在吃饭时吃掉您想吃的东西。＆＃39;，＆＃39;仍然需要保持身体活动。＆＃39 ;＆＃39;在第1行＆＃39;
这是查询：
if(isset($_POST['newdiet'])){
//deal with empty fields
$dietname = $_POST['dietname'];
$diettype = $_POST['diettype'];
$blurb = $_POST['blurb'];
$pro1 = $_POST['pro1'];
$pro2 = $_POST['pro2'];
$pro3 = $_POST['pro3'];
$con1 = $_POST['con1'];
$con2 = $_POST['con2'];
$con3 = $_POST['con3'];
$how = $_POST['how'];
$what = $_POST['what'];
$altcontent = $_POST['altcontent'];
$ing1 = $_POST['ing1'];
$ing2 = $_POST['ing2'];
$ing3 = $_POST['ing3'];
$ing4 = $_POST['ing4'];
$alt1 = $_POST['alt1'];
$alt2 = $_POST['alt2'];
$alt3 = $_POST['alt3'];
$duration = $_POST['duration'];
$imgurl = $_POST['imgurl'];
$newDiet = "INSERT INTO diets (dietname, diettype, blurb, pro1, pro2, pro3, con1, con2, con3, ing1, ing2, ing3, ing4, alt1, alt2, alt3, duration, how, what, alternatives, imgurl) VALUE ('$dietname', '$diettype','$blurb', '$pro1', '$pro2', '$pro3','$con1', '$con2', '$con3', '$ing1', '$ing2', '$ing3', '$ing4', '$alt1','$alt2', '$alt3', '$duration', '$how', '$what', '$altcontent', '$imgurl') ";
$newDietQry = mysqli_query($connect, $newDiet);
if(!$newDietQry){
die(mysqli_error($connect));
}elseif($newDietQry) {
$message3 = "New Diet Uploaded";
mail('emial@email.com', 'new diet uploaded', 'someone has uploaded a new diet to your website');
}
任何想法？我已经检查了网络错误日志，但它没有显示任何内容，只是500错误（！）。请帮帮我!!!!