Question

所以我试图在地址栏中显示ID。在以前的页面中，它已经可以工作了，但是在这一页面中，它似乎不起作用，我也不知道为什么。箭头显示了问题所在

我不希望看到的结果是该地址栏https://nottherealsiteurl/upload.php?id=

变成这个https://nottherealsiteurl/upload.php?id= $ row ['id']（数据库的ID）

<?php
$sql = "SELECT chauffeurs_naam, ritten.id ,chauffeurs.cc, ritten_date, ritten_totaal, ritten_naam, ritten_start, ritten_end, ritten_pauze, Kenteken, km_end, Onderhoudsrit
FROM ritten
JOIN chauffeurs
ON ritten.rit_cc = chauffeurs.cc
JOIN users
ON users.cc = chauffeurs.cc
WHERE users.id =".$_GET['id']. "";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
?>



<form action="upload.php?id=<?php echo $row['id'];?>" method="post" enctype="multipart/form-data">
<h3>Select image to upload:</h3>
    <input class="button button2" type="file" name="fileToUpload" id="fileToUpload" required="required">
    <input class="button button2" type="submit" value="Upload Image" name="submit">
</form>

在确实显示它的其他页面上。

它像这样放。再次，我把箭头放到它起作用的地方

$sql = "SELECT chauffeurs_naam, c.id, c.cc, c.chauffeurs_foto FROM chauffeurs c JOIN users u ON c.cc=u.cc WHERE u.id =".$_GET['id']."";
        $result = $conn->query($sql);
        if ($result->num_rows > 0) {
        while($row = $result->fetch_assoc()) {
        echo"<div class='col-xl-3 col-md-6 card'>";
        if (isset($row['chauffeurs_foto'])) {
          echo "<div class='caption'><img class='avatar-cards' alt='Generic placeholder thumbnail' src=''/></div>";
        }
        else {
          echo "<img class='avatar-cards' alt='Generic placeholder thumbnail' src='images/Test_Foto_Chauffeur.png'/>";
        }
        echo "<a href='Update_image.php?id=". $row['id'] ."'>Aanpassen</a>";
        echo "<div class='card-body'>";
        echo "<h4>". $row['chauffeurs_naam'] ."</h4>";
        echo"<p class='card-text'>";
        echo "Chauffeurs-nummer: ". -------------------->$row['id']<--------------------------- ."<br/>";
        echo "</p>";
        echo "</div>";
        echo "</div>";
        }

Answer 1

在此行upload.php？id =您缺少回声和半冒号

upload.php?id=<?php echo $row['id'];?>

尝试这个希望能奏效的

我已经运行了您的代码，并交叉检查了它的工作原理，一旦检查了简单的查询，如果工作了，就交叉检查了您的查询是否获取了数据

<?php
$sql = "SELECT * from tab_1";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
?>



<html>
<head>
<link rel="stylesheet" href="styling.css">
<title>Image Upload Tutorial</title>
</head>
<body>
<center>
<h1>Php Photo Upload Tutorial</h1>
<form action="upload.php?id=<?php echo $row['id'];?>" method="post" 
enctype="multipart/form-data">
<h3>Select image to upload:</h3>
<input class="button button2" type="file" name="fileToUpload" 
id="fileToUpload" required="required">
<input class="button button2" type="submit" value="Upload Image" 
name="submit">
</form>
</center>
</body>
</html>

完整的代码可以正常工作

单击链接后，php行ID不会显示在adressbar中

1 个答案: