在图像更新过程中，除了我更新的图像外，其他图像都是空的。

Question

当其他图像被更新时，我尝试了很多不同的方法来保留这些图像。一切都很好，只是当我上传或更新其中一个图像时，其他图像被删除或空保存到数据库中。

<?php
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
session_start();
    include ("connection.php");
    include ("functions.php");
    $user_data = check_login($con);

if(count($_POST)>0) {

if(!empty($_FILES['garden_photo1']['name'] . $_FILES['garden_photo2']['name']) && isset($_FILES["garden_photo1"]["name"], $_FILES['garden_photo2']['name'] )){

$id =  $_POST['id'];
$cultivar_name = $_POST['cultivar_name'];    
$breeder = $_POST['breeder'];
$lineage1 = $_POST['lineage1'];
$garden_photo1 = $_FILES['garden_photo1']['name'];
$ImageName = $_FILES['garden_photo1']['name'];
$ImageName2 = $_FILES['garden_photo2']['name'];
$target = "/cultivar/images/" . $ImageName;
$target2 = "/cultivar/images/" . $ImageName2;
move_uploaded_file($_FILES['garden_photo1']['tmp_name'], $target);
move_uploaded_file($_FILES['garden_photo2']['tmp_name'], $target2);
$stmt = $con->prepare("UPDATE cultivar_db set  cultivar_name = ?, breeder = ?, lineage1 = ?, garden_photo1 = ?,  garden_photo2 = ?  WHERE id= ? ");
$stmt->bind_param( "sssssi",  $cultivar_name, $breeder, $lineage1, $garden_photo1, $garden_photo2, $id);
$stmt->execute();
  }else {
$id =  $_POST['id'];
$cultivar_name = $_POST['cultivar_name'];    
$breeder = $_POST['breeder'];
$lineage1 = $_POST['lineage1'];
$stmt = $con->prepare("UPDATE cultivar_db set  cultivar_name = ?, breeder = ?, lineage1 = ?  WHERE id= ? ");
$stmt->bind_param( "sssi",  $cultivar_name, $breeder, $lineage1,  $id);
$stmt->execute();
  }
$message = "Cultivar Record Modified Successfully!";
}



$id =  $_GET['id'];   
$sql = "SELECT * FROM cultivar_db WHERE id=?"; 
$stmt = $con->prepare($sql); 
$stmt->bind_param("i", $id);
$stmt->execute();
$result = $stmt->get_result(); 
$row = mysqli_fetch_array($result);

?>

<html>
<head>
<title>Update Cultivar Data</title>
<style>
.success{color: green; font-size:20px; fonte-weight:bold;}

</style>
</head>
<body>
<form name="cultivar_db" method="post"  enctype="multipart/form-data">


<div style="padding-bottom:5px;">
<p><a href="cultivar-database.php">Go Back to Cultivar Database </a>&nbsp;&nbsp;>&nbsp;&nbsp;Edit <?php echo $row["cultivar_name"];?>  Page</p>
</div>
 <br>
<input type="hidden" name="id" value="<?php echo $row['id']; ?>" >
<br>
Cultivar Name: <br>
<input  name="cultivar_name"  value="<?php echo $row['cultivar_name']; ?>">
<br>
Breeder: <br>
<input type="text" name="breeder" value="<?php echo $row['breeder']; ?>">
<br>
Lineage 1 :<br>
<input type="text" name="lineage1"  value="<?php echo $row['lineage1']; ?>">
<br>
Garden Photo 1: <img id="output" src="../cultivar/images/<?php echo $row["garden_photo1"];?>" style="height:auto; width:70px;">
<br>
<br>
<input id="garden_photo1" type="file" name="garden_photo1"  >

<br>
Garden Photo 1: <img id="output" src="../cultivar/images/<?php echo $row["garden_photo2"];?>" style="height:auto; width:70px;">
<br>
<br>
<input id="garden_photo1" type="file" name="garden_photo2" >

<br>
<br>
<input id="submit" type="submit" name="submit" value="Submit" >
<br>
<br>
<div class="success">
<?php if(isset($message)) { echo $message; } ?>
</div>

</form>

</body>
</html>

Answer 1

现在，你似乎只是在区分这两种情况，根本没有上传任何图像，或者上传了两张图片。您需要处理的情况是，其中一个图片是上传的，而另一个没有，以及。

但是，与其为所有四种可能的组合编写不同的分支(没有上载图像；图像1被上传但图像2没有上载；图像1没有上载但图像2已经上载；或者两者都被上载)，如果您单独处理每个图像上传，则会更有意义：

图片1上传了吗？然后处理上传，并更新数据库中的图像1的列。

图片2上传了吗？然后处理上传，并更新数据库中的图像2的列。