我知道我有两个foreach但是当我将一个txt的glob分配给var时它不起作用。它应该为每个html文件找到写对应的txt文件等等,只有一次。我怎么能动摇这个?
<?php
echo '<h2>Escolha o Esquema de MetaDados a utilizar</h2>';
echo '<table cellspacing="0" cellpadding="0" border="0" class="table">';
echo '<tr>';
echo '<th>Esquema de MetaDados</th>';
echo '<th>Descrição</th>';
echo '</tr>';
foreach (glob("Templates/Dinamico/*.html") as $filename) {
foreach (glob("Templates/Dinamico/*.txt") as $txtname) {
$fileH = basename($filename, ".html");
$fileT = basename($txtname, ".txt");
if ($fileH = $fileT) {
$txt = file_get_contents($txtname);
if ($filename != "Templates/Dinamico/formD1.html") {
echo '<tr>';
echo '<td>';
echo "<a strong href='" . $filename . "'>" . $fileH . "</a>";
echo '</td>';
echo '<td>';
echo $txt;
echo '</td>';
echo '</tr>';
//just checking the files
echo "$filename size " . filesize($filename) . "\n";
}
}
}
}
答案 0 :(得分:0)
<?php
echo '<h2>Escolha o Esquema de MetaDados a utilizar</h2>';
echo '<table cellspacing="0" cellpadding="0" border="0" class="table">';
echo '<tr>';
echo '<th>Esquema de MetaDados</th>';
echo '<th>Descrição</th>';
echo '</tr>';
foreach (glob("Templates/Dinamico/*.html") as $filename) {
foreach(glob("Templates/Dinamico/*.txt") as $txtname){
$fileH= basename($filename,".html");
$fileT= basename($txtname,".txt");
if($fileH==$fileT){
$txt= file_get_contents ( $txtname);
if($filename != "Templates/Dinamico/formD1.html"){
echo'<tr>';
echo'<td>';
echo "<a name='".$fileH."'strong href='Templates/Dinamico/formD1.php?id=$fileH'>".$fileH."</a>";
echo'</td>';
echo'<td>';
echo $txt;
echo'</td>';
echo'</tr>';
}
}
}
}
?>