为什么错误信息不显示？

Question

我对 PHP 非常陌生（自 9 月以来才开始这样做，所以如果这似乎是一个愚蠢的问题，我深表歉意，我非常困惑，无法找到答案！）并且无法弄清楚为什么我的错误消息表单为空时用户提交表单时不显示。

这是我的代码：

    <?php
$salonid = "";
    if (!$db_server){
            die("Unable to connect to MySQL: " . mysqli_connect_error($db_server));
            $db_status = "not connected";
    }else{
        //Capture form data, if anything was submitted
        if (isset($_GET['salonid']) and ($_GET['salonid'] != '')){
            $salonid = clean_string($db_server, $_GET['salonid']);
            //If connected, get Salons from database and write out
            mysqli_select_db($db_server, $db_database);
            $query = "SELECT ID, salon_name, address, postcode, telephone, email, website FROM salon WHERE ID=$salonid";
            $result = mysqli_query($db_server, $query); 
            if (!$result) die("Query failed: " . mysqli_error($db_server));

            while($row = mysqli_fetch_array($result)){
                $str_result .= "<h2>" . $row[ 'salon_name'] . "</h2>";
                $str_result .= "<p>" . $row['address'] . "</p>";
                $str_result .= "<p>" . $row['postcode'] . "</p>";
                $str_result .= "<p>" . $row['telephone'] . "</p>";
                $str_result .= "<p>" . $row['email'] . "</p>";
                $str_result .= "<p>" . $row['website'] . "</p>";

            }
            mysqli_free_result($result);
        }else{
            $str_result = "<h2>No salon selected</h2>";

        }

    }
    echo $str_result;
?>

<?php 

if(trim($_POST['submit']) == "Submit comment"){

    //Get any submitted comments and insert
    $comment = clean_string($db_server, $_POST['comment']);
    if ($comment != '') {
        $name=$_FILES['photo']['name'];
        if ($name = "") $error .= "<p class='error'>You must upload an image!</p>";
        $originalname=$_FILES['photo']['name'];
        $type=$_FILES['photo']['type'];
        if ($type=="image/jpeg")  $type=".jpeg"; //if true change 
        else if ($type=="image/jpg") $type=".jpg";// if not true check this one
        else if ($type=="image/png") $type=".png";
        $name=uniqid() . $type;

        $path="images/" . $name;
        $tempname=$_FILES['photo']['tmp_name'];
        $size=$_FILES['photo']['size'];
        //Error checking
        if ($size >1000000) $error .= "<p class='error'>Your image file is to big, it have to be less than 200 mb</p>";
        if ($error=="") {
            if (move_uploaded_file($tempname, $path)){

                $uploadquery="INSERT INTO comments (comment, imagename, salonID, userID) VALUES ('$comment', '$path', $salonid, ". $_SESSION['userID'].")";
                mysqli_query($db_server,$uploadquery) or die ("Insert failed " . mysqli_error($db_server) . " " . $uploadquery);
                $message= "<h2>Thanks for your comment!</h2><p>Your upload was succesful</p>";
            }
        }
    }

}

//Print out existing comment
$query = "SELECT * FROM comments JOIN users ON comments.userID = users.ID WHERE salonID=$salonid"; 
$result = mysqli_query($db_server, $query);
if (!$result) die("Database access failed: " . mysqli_error($db_server));
while ($row = mysqli_fetch_array($result)){
        $str_comments .="<h2>" . $row['Username'] ."</h2>";
        $str_comments .= "<p>" . $row['comment'] . "</p>";
        $str_comments .="<img src='" . $row['imagename'] ."' />";
}

mysqli_free_result($result);

?>  
<div id="form">
<table><form id='review' action='salonpage.php?salonid=<?php echo $salonid; ?>' method='post' enctype='multipart/form-data'>
<th><h2> Do you want to review the service you recieved?</h2></th>
<tr><td><textarea name="comment" rows="6" cols="40">Write something here!</textarea></td></tr>
<tr><td><input type='file' name='photo' accept='image/jpg, image/jpeg, image/png'/></td></tr>
<br/>
<tr><td><input type='submit' id='submit' name='submit' value='Submit comment' /></td></tr>
</form></table>
<?php echo $message;
    echo $str_comments; ?>
</div>
<?php mysqli_close($db_server); ?>

Answer 1

我认为，您在谈论 $error，如果我是正确的，那么您没有在上面提到的代码中回显 $error 变量。还有一件事在 if 块“if ($comment != '')”上添加 else 部分 else {$error.="no comment entered" }

Answer 2

if ($comment != '') {
        $name=$_FILES['photo']['name'];
        if ($name = "") $error .= "<p class='error'>You must upload an image!</p>";
        $originalname=$_FILES['photo']['name'];

在这段代码中你正在使用

$name = ""

这是一个赋值运算符，您需要在 if 条件中使用比较运算符==或===