doudouba4520 2015-07-20 14:14
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too long

I have 10 rows, however when I do the following, it's only giving me the last row. Any kind of help I can get on this is great appreciated!

$query = "select * FROM `$table`.`channels` WHERE `country`='vietnam' ORDER BY `chanid`"; 
  $result = mysql_query($query,$db) or die(mysql_error());
  $data = array();
  while($row = mysql_fetch_array($result)) {
      $chanid = $row['chanid'];

        $data[navtitle] = "$chanid - $row[title]";
        $data[navurl] = "http://www.localhost.com/vietnam.php?chanid=$row[chanid]&country=$row[country]";
        $data[vid_art] = "$chanart";

  }

$array2=array_merge(array($array,array($data));

$JSON=json_encode($array2);

echo $JSON;

My $data array is only outputs the last row of my mysql fetch. How can I get it to pull out all 10 rows that I have?

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1条回答 默认 最新

  • douna2633 2015-07-20 14:21
    关注

    Each time you go through the

    while($row = mysql_fetch_array($result)) {
    

    loop, you're setting $data to a new value.

    This means it'll only ever contain the last row's worth of data.

    Your code should look something like this.

    $array2 = array();
    while ($row = mysql_fetch_array($result)) {
        $data = array();
        $chanid = $row['chanid'];
    
        $data['navtitle'] = "$chanid - $row[title]";
        $data['navurl'] = "http://www.localhost.com/vietnam.php?chanid=$row[chanid]&country=$row[country]";
        $data['vid_art'] = $chanart;
    
        $array2[] = $data;
    }
    
    $JSON = json_encode($array2);
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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