我有一个功能,为页面标题创建独特的slug.它会检查页面表中的slug是否可用,然后通过相应地添加’-int’来创建一个独特的slug.该函数对前三个条目工作正常,例如,三次输入的’test slug’将创建’test-slug-1′,’test-slug-2’和’test-slug-3′.然后,我得到第四个条目的错误“致命错误：超过30秒的最大执行时间”.逻辑应该有一些问题,任何人都可以帮我找到它.Below是代码：
function createSlug($title,$table_name,$field_name) {
global $db_connect;
$slug = preg_replace("/-$/","",preg_replace('/[^a-z0-9]+/i',"-",strtolower($title)));
$counter = 1;
do{
$query = "SELECT * FROM $table_name WHERE $field_name = '".$slug."'";
$result = MysqLi_query($db_connect,$query) or die(MysqLi_error($db_connect));
if(MysqLi_num_rows($result) > 0){
$count = strrchr($slug,"-");
$count = str_replace("-",$count);
if($count > 0){
$length = count($count) + 1;
$newSlug = str_replace(strrchr($slug,"-"),'',$slug);
$slug = $newSlug.'-'.$length;
$count++;
}else{
$slug = $slug.'-'.$counter;
}
}
$counter++;
$row = MysqLi_fetch_assoc($result);
}while(MysqLi_num_rows($result) > 0);
return $slug;
}
只需使用一个查询即可为您完成所有繁重的工作……
原文链接：https://www.f2er.com/php/137156.html$slug = preg_replace("/-$/",strtolower($title)));
$query = "SELECT COUNT(*) AS NumHits FROM $table_name WHERE $field_name LIKE '$slug%'";
$result = MysqLi_query($db_connect,$query) or die(MysqLi_error($db_connect));
$row = $result->fetch_assoc();
$numHits = $row['NumHits'];
return ($numHits > 0) ? ($slug . '-' . $numHits) : $slug;
