选择包含特定值

时间:2016-09-10 19:02:58

标签: php mysql

我有一个Sql表,我需要搜索members列,以查看是否有任何值与之前获得的值相同(例如:76561198119598543),如果是,那么我想要获取存储在同一行的name下的值......

下面的链接是表格的示例以及如何设置成员列的示例。 http://nbdcommunity.com/staffimages/example.png

我不是一个博士学位的人,并且已经尝试了我能找到的所有东西...... $gangsql = 'SELECT members, name, owner FROM 'gangs' WHERE owner LIKE "$playerid%"';(playerid是我想要比较的值)

3 个答案:

答案 0 :(得分:0)

列成员中的值看起来不是数字,而是字符串。假设情况如此,您可以尝试:

throw new InvalidArgumentException('Cannot end a section without first starting one.');

这将返回" name"的值对于"成员"的所有行列包含您检查的号码。

答案 1 :(得分:0)

以下是我对您的问题的理解:
它就像一个搜索框。如果要查看的数据匹配,则显示它?

示例:

PHP文件

    func collectionView(collectionView: UICollectionView, layout collectionViewLayout: UICollectionViewLayout, sizeForItemAtIndexPath indexPath: NSIndexPath) -> CGSize {

    if (indexPath.row < myComments!.count) {

        if let messageText = myComments![indexPath.row].text  {
            let size = CGSizeMake(self.view.frame.width / 2, 100)
            let options = NSStringDrawingOptions.UsesFontLeading.union(.UsesLineFragmentOrigin)
            let estimatedFrame = NSString(string: messageText).boundingRectWithSize(size, options: options, attributes: [NSFontAttributeName: UIFont.systemFontOfSize(14)], context: nil)
            return CGSizeMake(view.frame.width, estimatedFrame.height + 50)
        }
    }
    return CGSizeMake(view.frame.width, 100)

}

HTML HERE

    <?php

    //Connection here

    if isset($_POST["member"]){
    $get_data = $_POST["member"];
    //Display matched Data
    $sql = mysql_query("SELECT * FROM table_name WHERE member = '$get_data' ");
    while($row = mysql_fetch_array($sql)){
    //Declare variables
    $this_name = $row["name"];

    //Display Variable
    echo 
    'The name you are looking for is this person; on this case it\'s a number <u>'.$this_name.'</u>';

    }
    //END WHILE LOOP
    }else{
    //No match

     echo '
     <script>
     alert("Data has no match. Please try again");
     window.location= "php_page_name.php";
     </script>
         ';
     }

     ?>

希望这会有所帮助。祝你好运!

答案 2 :(得分:0)

$lastValue = "76561198119598543";

$c = new mysqli($server, $user, $password) or die("cant connect");

if($res = $c->query("SELECT name FROM table WHERE members='$lastValue'")){
  if($res->num_rows > 0){
     $resAssoc = $res->fetch_assoc();
     $name = $resAssoc["name"];
  }
} else {
   die($c->error);
}

$c->close();