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Php 获取链接和图像的多个foreach_Php - Fatal编程技术网

Php 获取链接和图像的多个foreach

Php 获取链接和图像的多个foreach,php,Php,我想从两个文本文件中获取图像url和图像链接 每人两个 $files = basename($_SERVER["SCRIPT_FILENAME"], '.php') ; $files = preg_replace("/.+?(\\d+).*/", "$1", $files); $count2 = $files; $page2 = $count2 ; $file2 = 'link.txt'; $per_page2 = 4; $content2 = file_get_contents($file

我想从两个文本文件中获取图像url和图像链接
每人两个

$files = basename($_SERVER["SCRIPT_FILENAME"], '.php') ;

$files = preg_replace("/.+?(\\d+).*/", "$1", $files);

$count2 = $files;
$page2 = $count2 ;
$file2 = 'link.txt';
$per_page2 = 4;
$content2 = file_get_contents($file2);
$array2 = explode(PHP_EOL, $content2);
$start2 = --$page2 * $per_page2;
$lines2 = array_slice($array2, $start2, $per_page2);
$output2 = '';


foreach ($lines2 as $line2)
{
    $count = $files;
    $page1 = $count ;
    $file = 'linkimg.txt';
    $per_page = 4;
    $content = file_get_contents($file);
    $array = explode(PHP_EOL, $content);
    $start = --$page1 * $per_page;
    $lines = array_slice($array, $start, $per_page);
    $output = '';

    foreach ($lines as $line)
    {
        echo '<div class="li4">';

        echo '<a href="' .$line2 .'">'  .'<img src="'   .$line .'">' .'</a>';

        echo '</div>';
    }
}
$files=basename($\服务器[“脚本\文件名”],'.php');
$files=preg_replace(“/。+?(\\d+).*/”、“$1”、$files);
$count2=$files;
$page2=$count2;
$file2='link.txt';
$per_page2=4;
$content2=文件获取内容($file2);
$array2=分解(PHP_EOL,$content2);
$start2=--$page2*$per_page2;
$lines2=阵列片($array2,$start2,$per_page2);
$output2='';
foreach($lines2作为$line2)
{
$count=$files;
$page1=$count;
$file='linkimg.txt';
每页$4;
$content=file\u get\u contents($file);
$array=explode(PHP\u EOL,$content);
$start=--$page1*$每页;
$lines=数组\u切片($array,$start,$per\u页);
$output='';
foreach($line作为$line)
{
回声';
回声';
回声';
}
}
我想显示图像的链接到其他链接,但不工作 可以帮助我。

请尝试以下代码

在此场景中使用for循环

假设: 1.两个文件有相等的行

$files = basename($_SERVER["SCRIPT_FILENAME"], '.php') ;

$files = preg_replace("/.+?(\\d+).*/", "$1", $files);

$count2 = $files;
$page2 = $count2 ;
$file2 = 'link.txt';
$per_page2 = 4;
$content2 = file_get_contents($file2);
$array2 = explode(PHP_EOL, $content2);
$start2 = --$page2 * $per_page2;
$lines2 = array_slice($array2, $start2, $per_page2);
$output2 = '';


 $count = $files;
    $page1 = $count ;
    $file = 'linkimg.txt';
    $per_page = 4;
    $content = file_get_contents($file);
    $array = explode(PHP_EOL, $content);
    $start = --$page1 * $per_page;
    $lines = array_slice($array, $start, $per_page);
    $output = '';

for ($i=0;$i<count($lines2);$i++)
{
        echo '<div class="li4">';

        echo '<a href="' .$line2[$i] .'">'  .'<img src="'   .$line[$i] .'">' .'</a>';

        echo '</div>';

}
$files=basename($\服务器[“脚本\文件名”],'.php');
$files=preg_replace(“/。+?(\\d+).*/”、“$1”、$files);
$count2=$files;
$page2=$count2;
$file2='link.txt';
$per_page2=4;
$content2=文件获取内容($file2);
$array2=分解(PHP_EOL,$content2);
$start2=--$page2*$per_page2;
$lines2=阵列片($array2,$start2,$per_page2);
$output2='';
$count=$files;
$page1=$count;
$file='linkimg.txt';
每页$4;
$content=file\u get\u contents($file);
$array=explode(PHP\u EOL,$content);
$start=--$page1*$每页;
$lines=数组\u切片($array,$start,$per\u页);
$output='';

对于($i=0;$i您在这里想要实现什么
files=basename($\u SERVER[“SCRIPT\u FILENAME”],'.php');$files=preg\u replace(“/.+?(\\d+).*/”,“$1”,$files);
?是否向脚本传递参数?我这样问是因为
$files
的值被传递,然后被视为一个整数。如果使用脚本名称传递值,则不正确。请使用参数运行脚本。;$files输出是页码,例如:如果页面加载=2,则脚本获取链接url行5,6,7,8。如果页面加载=3,脚本获取第9,10,11,12行,这实际上是一个仅代码(低质量)的答案,对于不清楚的问题有语法错误。请删除此答案。OP发布完整/清晰/可回答的问题后,您可以选择取消删除或发布新答案。这项工作,当我更改$line[i],$line2[i]至$lines[$i],$lines2[$i]…谢谢