【问题标题】:Php file_get_content not working to fetch local imagephp file_get_content 无法获取本地图像
【发布时间】:2020-06-04 09:43:03
【问题描述】:

我正在尝试使用 file_get_contents() 将图像转换为字符串到 encode_Base64,但它显示文件未找到

<?php
$conn = new mysqli("localhost","root","password","table");

$hotel = $_GET['hotel'].'menu';


$query="SELECT * FROM $hotel";

if(($result = $conn->query($query))==TRUE){
    $product = array();
    while($row = $result->fetch_assoc()){
        $data= base64_encode(file_get_contents("Images\\".$row['image'].".jpg"));
        echo $data;
        $item = array(
            "id" => $row['id'],
            "name" => $row['Name'],
            "price" => $row['Price'],
            "description"  => $row['Description'],
            "img" => $data);
        array_push($product , $item);
        break;
    }
    echo "Images\\".$row['image'].".jpg";
    echo "<image src='Images\\".$row['image'].".jpg'>";
    //echo json_encode($product);
}

?>

但在最后一行,当我回显相同的路径时,它会显示图像

那我该怎么办

【问题讨论】:

  • 据我所知,您不能使用 prepared statements 作为表名。更好的选择是拥有白名单并将您的表名包装在 ```

标签: php file-get-contents


【解决方案1】:

您遗漏了部分代码。 它会返回你想要的

<?php
$conn = new mysqli("localhost","root","password","table");

$hotel = $_GET['hotel'].'menu';


$query = "SELECT * FROM `$hotel`";

if (($result = $conn->query($query)) == TRUE) {
    $product = array();
    while ($row = $result->fetch_assoc()) {
        //changed
        $path = "Images\\" . $row['image'] . ".jpg";
        $data = 'data:image/jpg;base64,' . base64_encode(file_get_contents($path));

        echo $data;
        $item = array(
            "id" => $row['id'],
            "name" => $row['Name'],
            "price" => $row['Price'],
            "description" => $row['Description'],
            "img" => $data);
        array_push($product, $item);
        break;
    }
    echo "Images\\" . $row['image'] . ".jpg";
    //echo "<image src='Images\\".$row['image'].".jpg'>";
    echo '<image src="' . $data . '">';
    echo '<image src="' . $path . '">';
    //echo json_encode($product);
}

【讨论】:

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