我的mysql表中有三个归档:id,url,status
如何检查列url中的所有网址并将1(可用)或0(不可用)写入状态列?
要在php w / o mysql中手动检查url我可以使用:
<?php
function Visit($url){
$agent = "Mozilla/4.0 (compatible; MSIE 5.01; Windows NT 5.0)";
$ch = curl_init();
curl_setopt ($ch, CURLOPT_URL,$url );
curl_setopt($ch, CURLOPT_USERAGENT, $agent);
curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt ($ch,CURLOPT_VERBOSE,false);
curl_setopt($ch, CURLOPT_TIMEOUT, 5);
$page=curl_exec($ch);
//echo curl_error($ch);
$httpcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close($ch);
if($httpcode >= 200 && $httpcode < 300){
return true;
}
else {
return false;
}
}
if(Visit("http://www.google.com")){//maybe make it a variable from a result of a mysql select, but how to process it one by one?
echo "Website OK"; //maybe somesql here to wtite '1'
}
else{
echo "Website DOWN";//maybe somesql here to wtite '0'
}
?>
答案 0 :(得分:1)
这在性能方面不好,因为在循环中进行查询是糟糕的设计,您应该建立批量更新并最终运行查询,但我现在没有足够的时间来详细说明。这只是为了让你开始:
$sql = "SELECT id,url FROM mytable";
$res = mysql_query($sql) or die(mysql_error());
if($res)
{
while($row = mysql_fetch_assoc($res))
{
$status = visit($row['url']) ? '1' : '0';
$id = $row['id'];
$update = "UPDATE mytable SET status = $status WHERE id = $id";
$res = mysql_query($update) or die(mysql_error());
}
}
答案 1 :(得分:0)
echo mysql_num_rows($result) ? '1' : '0';
答案 2 :(得分:0)
您可以使用sql查询,例如:
select id, url from urltable;
update urltable set status=1 where id=99;
那么,如何在访问函数中使用此代码:
$link = mysql_connect('localhost', 'test', 'pppppp');
$db_selected = mysql_select_db('test', $link);
$query = "select id, url from urltable";
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
$visitid = $row['id'];
$visiturl = $row['url'];
$visitstatus = Visit($visiturl)? 1: 0;
$upquery = sprintf("update urltable set status=%d where id=%d", $visitstatus, $visitid);
$upresult = mysql_query($upquery);
}