如何打印包含preg_match中单词的行？

Question

A field in mysql table "server_var_dump" which contains many details including USER_AGENT of the visitors of site. Now I only need USER_AGENTs containing line. Now I have written following script to match the USER_AGENT of the output string. Now I need to print only those lines which contains the USER_AGENT

$result = mysql_query("SELECT server_var_dump FROM pageviews LIMIT 0,10");

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    //printf("%s", $row["server_var_dump"]);
    if(preg_match("/[USER_AGENT]/", $row["server_var_dump"])){
    echo "match found.";
    }
    else
    echo "No matches found";     
}

Please suggest me how do I print the line which contains USER_AGENT?

Thanks!

Answer 1

What's wrong with just printing the row once you got it?

$match_found = false;
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    if(preg_match("/[USER_AGENT]/", $row["server_var_dump"])){
        echo $row["server_var_dump"];
        $match_found = true;
        break;
    }
}
if(!$match_found) {
    echo "No matches found";
}

Alternatively, you'll need to get the matched values like so:

$match_found = false;
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    $matches = array();
    if(preg_match("/[USER_AGENT](.+)/", $row["server_var_dump"], $matches)){
        echo $matches[1];
        $match_found = true;
    }
}
if(!$match_found) {
    echo "No matches found";
}