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我试图从两个不同的文本文件中获取两个不同的数字，将它们放入一个变量中，并将其与 MYSQL 数组进行比较。我的问题是，当我定义三个变量时：

$num = 42;

$whichPlaylist = 2;

$joint = "$whichPlaylist $num"

并使用数组进行测试，它可以工作。但是，如果我从文本文件中取出 2 和 42，它就不起作用。我试过使用 $trim 但这似乎并没有解决它。为什么当我定义我的变量时它起作用，但当我从文本文件中获取它们时它不起作用？他们有完全相同的东西（虽然文本文件有一个新行）。

这是我的代码：

//Connect to DB
$config = array(
'host'       => 'localhost',
'username'   => '******',
'password'   => '******',
'dbname'     => '******'
);

$db = new PDO('mysql:host='.$config['host'].';dbname='.$config['dbname'],$config['username'],$config['password']);
$query = $db->query("SELECT `recentlyplayed`.`numplayed`, `recentlyplayed`.`id` FROM `recentlyplayed`");


//Put numbers from text files into variables
$num = file_get_contents('/home/*****/num.txt'); // has the value "42" with new line
$whichPlaylist = file_get_contents('/home/*****/whichplaylist.txt'); // has the value "2" with newline
$playlist3_num = file_get_contents('/home/*****/playlist3_num.txt');





//$whichPlaylist = 2;
//$num = 42;
$joint = "$whichPlaylist $num";
$trimmed = trim("$joint");
echo $trimmed;


while ($row = $query->fetch(PDO::FETCH_ASSOC)) {
if (in_array($trimmed, $row)){
    echo " In Array!"; //This does NOT work, but if I use the self-defined variables $whichPlaylist and $num (and comment out file_get_content variables) it DOES work.

}

}
4

2 回答 2

6

在加入之前，您需要从 file_get_contents 中修剪变量；否则换行符将在结果字符串中。修剪只需要字符串开头和结尾的空格，这会将它放在中间。

于 2013-01-10T00:32:40.950 回答
1

在加入它们之前，您需要修剪这些值。如：

$num = trim(file_get_contents('/home/*****/num.txt')); // has the value "42" with new line
$whichPlaylist = trim(file_get_contents('/home/*****/whichplaylist.txt')); // has the value "2" with newline
$playlist3_num = file_get_contents('/home/*****/playlist3_num.txt');

$joint = "$whichPlaylist $num";
于 2013-01-10T00:33:56.830 回答