php - BLOB 图像返回损坏的图像

Question

我在 SO 上找到了很多对其他人有效但对我无效的解决方案!正确方向的指导将不胜感激。

注意:我知道很多人可能不喜欢将图像存储在 blob 中，但对于我的特定应用程序来说，这就足够了。

上传页面上的 HTML 代码片段(在包含其他数据的表单中):

<div class="row">
            <div class="col-md-6">
                <div class="form-group">
                    <label>Summary Image</label>
                        <br /><input name="summaryImage" id="imgInp1" type="file" accept="image/*"  onchange="Prev1(this)" />
                        <br /><img id="imgPrev1" src="" alt="Summary Image Preview" style="width:99%; height: 400px; margin-top:10px;" />
                </div>
            </div>
            <div class="col-md-6">
                <div class="form-group">
                    <label>Article Image</label>
                        <br /><input name="articleImage" id="imgInp2" type="file" accept="image/*"  onchange="Prev2(this)" />
                        <br /><img id="imgPrev2" src="" alt="Article Image Preview" style="width:99%; height: 400px; margin-top:10px;" />
                </div>
            </div>
        </div>

上传页面上的 PHP 代码片段

//Summary Image
$summaryimage = $_FILES['summaryImage'];
if(is_uploaded_file($_FILES['summaryImage']['tmp_name']) && getimagesize($_FILES['summaryImage']['tmp_name']) != false)
{
    // Gather image info
    $size = getimagesize($_FILES['summaryImage']['tmp_name']);
    // Assign variables
    $type = $size['mime'];
    $imgfp = fopen($_FILES['summaryImage']['tmp_name'], 'rb');
    $size = $size[3];
    $name = $_FILES['summaryImage']['name'];
    $summaryimagesize = $_FILES['summaryImage']['size'];
    $maxsize = 15728640;


    // Check image file size
    if($summaryimagesize < $maxsize ) {
        //Image size checks out, insert blob into database
        mysql_query("UPDATE projects 
                        SET summaryimage='$summaryimage', summaryimagesize='$summaryimagesize' 
                    WHERE job='$job'") or die (mysql_error());;

        } else if($summaryimagesize > $maxsize) {
        // Image file size too big
        header("LOCATION: /projects.php?message=imagesize");
        }
    } else {
    // Any other error
    header("LOCATION: /projects.php?message=imageformat");
    }

    //Article Image
    $articleimage = $_FILES['articleImage'];
    if(is_uploaded_file($_FILES['articleImage']['tmp_name']) && 
        getimagesize($_FILES['articleImage']['tmp_name']) != false)
    {
        // Gather image info
        $size = getimagesize($_FILES['articleImage']['tmp_name']);
        // Assign variables
        $type = $size['mime'];
        $imgfp = fopen($_FILES['articleImage']['tmp_name'], 'rb');
        $size = $size[3];
        $name = $_FILES['articleImage']['name'];
        $articleimagesize = $_FILES['articleImage']['size'];
        $maxsize = 15728640;

        // Check image file size
        if($articleimagesize < $maxsize ) {
            //Image size checks out, insert blob into database
             mysql_query("UPDATE projects SET articleimage='$articleimage', articleimagesize='$articleimagesize' WHERE job='$job'") or die (mysql_error());;

            } else if($articleimagesize > $maxsize) {
            // Image file size too big
            header("LOCATION: /projects.php?message=imagesize");
            }
        } else {
        // Any other error
        header("LOCATION: /projects.php?message=imageformat");
        }

    header("LOCATION: /projects.php?message=success");

}

数据库将 blob 存储为 longblob 类型，没问题:

我用来编码和显示图像的代码:

<?php echo '<img src="data:image/jpeg;base64,' . base64_encode( $post['summaryimage'] ) . '" />'; ?>

页面显示的内容:

当我右键单击 > 复制图像位置时:

data:image/jpeg;base64,QXJyYXk=

Answer 1

QXJyYXk= 是 array 的 base64，因此您存储的是数组数据而不是图像。

解决方法是获取图像的内容并将其存储起来:

 $summaryimage = file_get_contents($_FILES['summaryImage']['tmp_name']);

因为您没有使用准备好的语句，所以您应该转义引号:

 $summaryimage = addslashes(file_get_contents($_FILES['summaryImage']['tmp_name']));

显示时你应该在编码之前再次删除它们:

base64_encode(stripslashes($post['summaryimage']))