从MySQL数据库中为longblob清空PHP输出

Question

我有一个Android应用程序拍摄照片，将位图转换为Base64并将Base64字符串提交到MySQL数据库（通过PHP）存储为longblob。这部分很棒！事实上，我可以从phpMyAdmin下载longblob作为完美的Base64字符串，并轻松转换为JPEG照片。

问题是我的PHP代码获取blob返回一个空字符串：

{
    "owner":"Unknown",
    "pet_name":"Unknown",
    "last_seen":"2019-04-09 11:17:19",
    "contact":"999-888-7654",
    "description":"rubber ducky, lotsa fun",
    ***"photo":""***,
    "location":"Some location"
}

PHP getter：

function getReports() {
    $stmt = $this->con->prepare("SELECT owner, pet_name, last_seen, contact, description, photo, location FROM Pets");
    $stmt->execute();
    $stmt->bind_result($owner, $pet_name, $last_seen, $contact, $description, $photo, $location);

    $reports = array();

    while($stmt->fetch()) {
        $report  = array();
        $report['owner'] = $owner;
        $report['pet_name'] = $pet_name;
        $report['last_seen'] = $last_seen;
        $report['contact'] = $contact;
        $report['description'] = $description;
        $report['photo'] = $photo;
        $report['location'] = $location;

        array_push($reports, $report);
    }

    return $reports;
}

一个有趣的旁注，如果不是上面的代码，我使用下面的代码，我得到完整的Base64字符串，但在整个过程中添加了escape（）和换行符（\ n）字符：

//Select everything from table
$sql= "SELECT * FROM Pets";

//Confirm results
if($result = mysqli_query($con, $sql)) {
    //Results? Create array for results and array for data
    $resultArray = array();
    $tempArray = array();

    //Loop through results
    while($row=$result->fetch_object()) {
        // Add each result in results array
        $tempArray=$row;
        array_push($resultArray,$tempArray);
    }

    //Encode array to JSON and output results
    echo json_encode($resultArray);
}

我想找到一种方法来修复上面的PHP代码。我想也许我的字符串太长了$photo值？任何建议都将非常感激。

更新：从Selecting Blob from MYSQL, getting null我确实管理到Base64现在输出而不是空字符串。但是，我仍然有转义和换行符的问题。

这里有什么帮助？

Answer 1

我设法得到原始函数通过强制转换输出Base64行：

    $stmt = $this->con->prepare("SELECT owner, pet_name, last_seen, contact, description, CAST(photo as CHAR(1000000) CHARACTER SET utf8) as photo, location FROM Pets");

虽然这允许接收Base64字符串，但它仍然包含不需要的字符。不需要的字符是由JSON_ENCODE引起的。以下是我用来解决它的问题。

基本上，1。删除添加的字符和2.告诉JSON_ENCODE不要使用JSON_UNESCAPED_SLASHES打印转义字符。

对于函数getReports（）

function getReports() {
    $stmt = $this->con->prepare("SELECT owner, pet_name, last_seen, contact, description, CAST(photo as CHAR(1000000) CHARACTER SET utf8) as photo, location FROM Pets");
    $stmt->execute();
    $stmt->bind_result($owner, $pet_name, $last_seen, $contact, $description, $photo, $location);

    $reports = array();

    while($stmt->fetch()) {
        $report  = array();
        $report['owner'] = $owner;
        $report['pet_name'] = $pet_name;
        $report['last_seen'] = $last_seen;
        $report['contact'] = $contact;
        $report['description'] = $description;
        $photo = str_replace("\n","",$photo);
        $photo = str_replace("\\/","/", $photo);
        $photo = stripcslashes($photo);
        $report['photo'] = $photo;
        $report['location'] = $location;

        array_push($reports, $report);
    }

    return $reports;
}

在Api脚本中，改变了返回

echo json_encode($resultArray);

至

echo json_encode($resultArray, JSON_UNESCAPED_SLASHES);

现在一切都很棒。这是最佳做法吗？我不确定。我肯定存储base64可能不是......