Question

大家好，我有一个代码，我必须选择一个图像，但如果我不想更新数据库中的现有图像怎么办？它是我系统中个人资料的更新。

我正在使用mysql数据库和mysqli编写的语句你能帮帮我吗？

这是我的工作代码。

<?php
$id_alum = $_GET['id'];
include('db/database_configuration.php');
if(ISSET($_POST['save'])){

    if(isset($_FILES['image']['name'])){
    $image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
    $image_name = addslashes($_FILES['image']['name']);
    $image_size = getimagesize($_FILES['image']['tmp_name']);
    move_uploaded_file($_FILES["image"]["tmp_name"], "featured_image/". $_FILES["image"]["name"]);
    $location = $_FILES["image"]["name"];
    //edit.....get input values
    if (empty($_POST['fullname'])){$fullname = 'NULL'; } else{ $fullname ="". mysqli_real_escape_string($conn, $_POST['fullname']) . "";}
    if (empty($_POST['job'])){$job_title = 'NULL'; } else{ $job_title ="". mysqli_real_escape_string($conn, $_POST['job']) . "";}
    if (empty($_POST['desc'])){$description = 'NULL'; } else{ $description ="". mysqli_real_escape_string($conn, $_POST['desc']) . "";}
    if (empty($_POST['desc2'])){$description2 = 'NULL'; } else{ $description2 ="". mysqli_real_escape_string($conn, $_POST['desc2']) . "";}

    $sql = "UPDATE `tblfeatured` SET image1=?, fullname=?, job_title=?, description=?, description2=? WHERE id_alum= $id_alum";

    $stmt = $conn->prepare($sql);

    // This assumes the date and account_id parameters are integers `d` and the rest are strings `s`
    // So that's 5 consecutive string params and then 4 integer params

    $stmt->bind_param('sssss', $location, $fullname, $job_title, $description, $description2);
    $stmt->execute();

    if ($stmt->errno) {
      echo "FAILURE!!! " . $stmt->error;
    }
    else {
        echo "<script>alert('Updated Successfully')</script>";
        echo '<script>window.location = "featured_result.php"</script>';
    }

    $stmt->close();
    }
}
?>

这是我的html文件

<form method = "POST" action = "update_featured_alumni.php?id=<?php echo $id_alum; ?>" enctype = "multipart/form-data">
        <label style="color: white; font-size: 12pt;">Change Image? Drag or click for an image</label>

        <div id="uploader" onclick="$('#photo').click()">
        <img src="" width="" />
        <div class="pull-right">Existing Image<img src="featured_image/<?php echo $image;?>" width="150" height="150"></div>
        </div>


        <input type="file" name="image"  id="photo"/>
        <div id = "file_name" style="color: white;"></div>
        <button class = "w3-btn w3-green w3-card-4" name = "save" title="Save"><font size="2"><span class = "fa fa-arrow-circle-o-down"></span> Save</font></button>

        <br>
        <font color="black">
        <input class = "form-control" type = "text" name= "fullname" placeholder = "Fullname" style="margin-bottom: 15px;" value="<?php echo $fname;?>">



        <input class = "form-control" type = "text" name= "job" placeholder = "Job Title & Workplace" style="margin-bottom: 15px;" value="<?php echo $job;?>">
        <textarea id="txtArea" name="desc" onkeyup="resizeTextarea('txtArea')" data-resizable="true" placeholder="Alumni Description"><?php echo str_replace('\r\n', "\r\n", $desc); ?></textarea>
        <textarea id="txtArea2" name="desc2" onkeyup="resizeTextarea('txtArea2')" data-resizable="true" placeholder="Alumni Description"><?php echo str_replace('\r\n', "\r\n", $desc2); ?></textarea>
        </font>
        <button class = "w3-btn w3-green w3-card-4" name = "save" title="Save"><font size="6"><span class = "fa fa-arrow-circle-o-down"></span> Save</font></button>
</form>

Answer 1

试试这个：

<?php
$id_alum = $_GET['id'];
include('db/database_configuration.php');
if(ISSET($_POST['save'])){
    $sql_img = "";

    if($_FILES['image']['size'] > 0){
      $image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
      $image_name = addslashes($_FILES['image']['name']);
      $image_size = getimagesize($_FILES['image']['tmp_name']);
      move_uploaded_file($_FILES["image"]["tmp_name"], "featured_image/". $_FILES["image"]["name"]);
      $location = $_FILES["image"]["name"];
      $location_db = addslashes($_FILES["image"]["name"]);
      $sql_img = "image1 = '".$location."', ";
    }
    //edit.....get input values
    if (empty($_POST['fullname'])){$fullname = 'NULL'; } else{ $fullname ="". mysqli_real_escape_string($conn, $_POST['fullname']) . "";}
    if (empty($_POST['job'])){$job_title = 'NULL'; } else{ $job_title ="". mysqli_real_escape_string($conn, $_POST['job']) . "";}
    if (empty($_POST['desc'])){$description = 'NULL'; } else{ $description ="". mysqli_real_escape_string($conn, $_POST['desc']) . "";}
    if (empty($_POST['desc2'])){$description2 = 'NULL'; } else{ $description2 ="". mysqli_real_escape_string($conn, $_POST['desc2']) . "";}
    $sql = "UPDATE `tblfeatured` SET ".$sql_img."fullname=?, job_title=?, description=?, description2=? WHERE id_alum= $id_alum";

    $stmt = $conn->prepare($sql);

    // This assumes the date and account_id parameters are integers `d` and the rest are strings `s`
    // So that's 5 consecutive string params and then 4 integer params

    $stmt->bind_param('ssss', $fullname, $job_title, $description, $description2);
    $stmt->execute();

    if ($stmt->errno) {
      echo "FAILURE!!! " . $stmt->error;
    }
    else {
        echo "<script>alert('Updated Successfully')</script>";
        echo '<script>window.location = "featured_result.php"</script>';
    }

    $stmt->close();
}
?>

用简单的话解释我添加的内容：$ sql_img最初设置为空。如果图像已过帐，则会更改为图像位置字符串。并添加到最终的sql脚本

如何从数据库更新数据但不更新php中的图像

1 个答案: