PHP数组，输出，仅输出类型='；图像/jpeg'；

我有php数组（$_FILE[fails]），我只需要输出千个名称，其中类型是“image/jpeg”或“image/jpg”

怎么做？ 用foreach

foreach ($_FILES["fails"] as $x => $y ) {
echo $y->....something here?
   if(($y->... == 'image/jpeg') AND (..)) {}
} 

谢谢大家!

试试看：

$output = array();
foreach ($_FILES["fails"]['type'] as $k => $type ) {
    if($type == 'image/jpeg' || $type == 'image/jpg')
        $output[] = $_FILES["fails"]['name'][$k];
}

for（$i=0；$i

这将回显数组中每个图像/jpg或图像/jpeg的名称和类型。

试试看

$allowTypes = array (
        "image/jpeg",
        "image/jpg" 
); // Ad More to list
$_FILE ['fails'] = array ();
foreach ( $_FILES ['image'] ['tmp_name'] as $key => $val ) {
    $fileName = $_FILES ['image'] ['name'] [$key];
    $fileType = $_FILES ['image'] ['type'] [$key];
    if ($_FILES ["image"] ["error"] [$key] != UPLOAD_ERR_OK) {
        $_FILE ['fails'] [] = $fileName;
        continue;
    }

    if (! in_array ( $fileType, $allowTypes )) {
        $_FILE ['fails'] [] = $fileName;
        continue;
    }
}

---

您不应该信任客户端发送给您的MIME类型。用于获得更可靠的答案，但即使如此，您仍然容易受到XSS的攻击。唯一可靠的清理方法是获取像素数据并创建新图像。

for ($i = 0; $i < count($_FILES['fails']['name']); $i++) {
    if ($_FILES['fails']['type'][$i] == "image/jpeg" || $_FILES['fails']['type'][$i] == "image/jpg") {
        echo "Name : " . $_FILES['fails']['name'][$i];
        echo "Type : " . $_FILES['fails']['type'][$i];
    }
}

$allowTypes = array (
        "image/jpeg",
        "image/jpg" 
); // Ad More to list
$_FILE ['fails'] = array ();
foreach ( $_FILES ['image'] ['tmp_name'] as $key => $val ) {
    $fileName = $_FILES ['image'] ['name'] [$key];
    $fileType = $_FILES ['image'] ['type'] [$key];
    if ($_FILES ["image"] ["error"] [$key] != UPLOAD_ERR_OK) {
        $_FILE ['fails'] [] = $fileName;
        continue;
    }

    if (! in_array ( $fileType, $allowTypes )) {
        $_FILE ['fails'] [] = $fileName;
        continue;
    }
}