简体   繁体   English

php array_merge

[英]php array_merge

I am getting an error on this line: 我在这条线上出现错误:

$ret=array_merge($ret,preg_ls($path."/".$e,$rec,$pat));

Error is: array_merge() Argument #2 is not an array 错误是:array_merge()参数2不是数组

I dont know know to solve this. 我不知道要解决这个问题。

Thank you. 谢谢。

function preg_ls($path=".", $rec=false, $pat="/.*/") {
    // it's going to be used repeatedly, ensure we compile it for speed.
    $pat=preg_replace("|(/.*/[^S]*)|s", "\\1S", $pat);
    //echo($pat);
    //Remove trailing slashes from path
    while (substr($path,-1,1)=="/") $path=substr($path,0,-1);
    //also, make sure that $path is a directory and repair any screwups
    if (!is_dir($path)) $path=dirname($path);
    //assert either truth or falsehoold of $rec, allow no scalars to mean truth
    if ($rec!==true) $rec=false;
    //get a directory handle
    $d=dir($path);
    //initialise the output array
    $ret=Array();
    //loop, reading until there's no more to read
    while (false!==($e=$d->read())) {
        //Ignore parent- and self-links
        if (($e==".")||($e=="..")) continue;
        //If we're working recursively and it's a directory, grab and merge
        if ($rec && is_dir($path."/".$e)) {
            $ret=array_merge($ret,preg_ls($path."/".$e,$rec,$pat));
            continue;
        }
        //If it don't match, exclude it
        if (!preg_match($pat,$e)) continue;
        //In all other cases, add it to the output array
        //echo($path."/".$e."<br/>");
        $ret[]=$path."/".$e;
    }
    //finally, return the array
    echo json_encode($ret);
}

An Array in PHP is not JSON. PHP中的Array不是JSON。 It's an array. 这是一个数组。 Simply return $ret; 只需return $ret;

You should return the array if you are expecting an array, not a string (as json_encode gives). 如果期望数组,而不是字符串(如json_encode给出),则应返回数组。

Also, you are using echo , not return . 另外,您正在使用echo ,而不是return echo prints to either stdout or the HTML body, depending on the PHP environment (though they are one and the same, just with redirects and a different environment to handle it). 根据PHP环境的不同, echo打印到stdout或HTML正文(尽管它们是相同的,只是带有重定向和不同的环境来处理它)。

return will cause a function to pass its return value to the caller, as expected (usually into a variable or another function); return将使函数按预期将其返回值传递给调用方(通常传递给变量或其他函数); without a return value, the function will always return NULL . 没有返回值,该函数将始终返回NULL

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM