Question

当我运行它时它只返回数据库中的1个结果但应该有2.你知道问题是什么或者你需要更多信息

示例数据库

URL             tags

www.google.com  google,search,information,cool,

www.youtube.com google,videos,



    $keywords=$_GET["Keywords"];                         
    $explodeKeywords = explode(" ", $keywords); 
    $currentWord = $explodeKeywords[$arrayKeywordCount];
    $arrayKeywordCount++;
    $URLSQUERY=mysql_query("Select URL 
            FROM tagslinks WHERE tags 
            LIKE '%$currentWord,%';");
    $URLSARRAY = mysql_fetch_array($URLSQUERY);

Answer 1

你必须在循环中获取所有结果，执行mysql_fetch_array一次，只得到1个结果。

for ($i = 0; $i < mysql_num_rows($URLSQUERY); $i++) {
    $URLSARRAY[] = mysql_fetch_array($URLSQUERY);
}

Answer 2

试试这个 - ＆gt;

$keywords=$_GET["Keywords"];                         
    $explodeKeywords = explode(" ", $keywords); 
    $currentWord = $explodeKeywords[$arrayKeywordCount];
    $arrayKeywordCount++;
    $URLSQUERY=mysql_query("Select URL 
            FROM tagslinks WHERE tags 
            LIKE '%$currentWord%'");
 while($res=mysql_fetch_array($URLSQUERY)){
  echo "<pre>";
  print_r($res);
}

喜欢MYSQL只获得1个结果

2 个答案: