想要替换谷歌索引我的链接，没有空格、逗号和其他字符
我的当前代码
<a href="report-<?php echo str_replace(" ", "-", $db['Subject'])?>-<?=$db['id']?>" class="read-more-button">Read More</a>
我正在寻找它为任何字符，如$，#，&不仅仅是空间。
<?php
$replace=str_replace(array(" ","'$","#","&","!",";"),'-',$db['Subject']);
echo "<a href='report-{$replace}-{$db['id']}' class='read-more-button'>Read More</a>";
?>
这将创建一个变量$replace，用str_replace'替换带有连字符的数组中的任何内容
这是我用来创建slugs 的
strtolower(preg_replace('/['s-]+/', '-', preg_replace('/[^A-Za-z0-9-]+/', '-', preg_replace('/[&]/', 'and', preg_replace('/['']/', '', trim($string))))));
str_replace函数允许您使用数组进行替换。像这样：
首先创建一个要替换的字符数组：
$replace = array(" ", "-", "_", "#", "+", "*");
然后在str_replace中给它一个数组名称：
$originalString = '<a href="http://www.somewhere.co.uk">My Link</a>';
$newString = str_replace($replace, "", $originalString);
$newString将删除$replace数组中的每个项。
您可以向str_replace函数发送array，如下所示：
$strip = array('%','$','#','&','!');
<a href="report-<?=str_replace($strip, '-', $db['Subject'])?>-<?=$db['id']?>" class="read-more-button">Read More</a>
然而，要创建URL，我使用以下内容：
<?php
function stripChars($str)
{
$bads = array(".","+"," ","#","?","!","&" ,"%",":","–","/","''","'","'"","”","“",",","£","’");
$goods = array("","-","-","-","" ,"" ,"and","" ,"" ,"" ,"","","","","","","","","","");
$str = str_replace($bads,$goods,$str);
return strtolower($str);
}
?>
<a href="<?=stripChars('report ' . $db['Subject'] . ' ' . $db['id'])?>" class="read-more-button">Read More</a>
来源于askingbox作者：
$f = 'fi?le.txt';
$f = str_replace(array('$', '#',' &', '!', '''','/',':','*','?','"','<','>','|'),' ',$f);
echo $f; // 'fi le.txt'
使用：
function make_safe($str){
return str_replace(array('$', '#',' &', '!', '''','/',':','*','?','"','<','>','|'),' ',$str);
}
<a href="report-<?php echo make_safe($db['Subject'])?>-<?=$db['id']?>" class="read-more-button">Read More</a>