我怎么能停止一个循环重复循环的PHP循环?

时间:2015-07-11 23:48:57

标签: php mysql polymer-1.0

我有一个聚合物(1.0)应用程序我正在尝试创建(基于拉伸教程联赛表),我无法弄清楚路由如果使用mvc,所以我选择了单页网站,引用不同的PHP文件。

我在MYSql中有一个表,我试图使用一个简单的echo表脚本合并到表中,但这会重复数百次。我怎么能阻止这个循环?

$result = mysql_query('select * from results',$connection) or die('cannot show tables');
while($tableName = mysql_fetch_row($result)) {

$sql = ('SELECT `home_team_name`,`away_team_name`,`home_goals_for`, `away_goals_for`, DATE_FORMAT(`fixture_date`, "%e %M %Y"), TIME_FORMAT(`fixture_time`, "%h:%i %p") FROM `results` ORDER BY fixture_date desc LIMIT 20  '.$table) or die('cannot show columns from '.$table);

echo '<h3>',$table,'</h3>';
$result2 = mysql_query($sql);
if(mysql_num_rows($result2)) {
    echo '<table cellpadding="4" cellspacing="0" class="table table-striped table-hover table-bordered">';
    echo '<tr class="info"><th>Home team</th><th>Away Team</th><th>Score</th><th>Score</th><th>Date<th>Time</th></tr>';
    while($row2 = mysql_fetch_row($result2)) {
        echo '<tr>';
        foreach($row2 as $key=>$value)  {
            echo '<td>',$value,'</td>';
        }
        echo '</tr>';
    }
    echo '</table><br />';
}
  }

我添加了&#39; &LT; 50&#39;但是这只返回50个表头行?

网站现在是http://jay-o.uk/#!/results css,其他数据现在还可以,只是这个讨厌的循环。

4 个答案:

答案 0 :(得分:0)

你需要设置一个&#34; LIMIT&#34;在第一个查询中,如果你想避免超时操作,我认为可以调用一个唯一的查询,返回你需要的所有信息,但我不知道,因为你的查询是无法理解的。

您的代码错误,第二个查询不使用null值的$ table变量以及将其放在limit参数之后的目的是什么?

$results = mysql_query('SELECT * FROM table_name ORDER BY field_name LIMIT 50');
if(mysql_num_rows($results)) {
  print '<table>';
  while ($row = mysql_fetch_row($results)) {
    print '<tr>';
    foreach ($row as $field_name => $field_value) print '<td>'.$field_value.'</td>';
    print '</tr>'
  }
  print '</table>';
}

答案 1 :(得分:0)

对不起,代码有点凌乱,我一直在研究这个问题,如果我感到沮丧,我不知道我哪里出错,如果我删除$ tableName变量我得到一个空数组,无论我是什么这是结果。

我在json尝试过无济于事......

$db = mysql_connect($host,$user,$pass);

if (!$db) {

    die('Could not connect to db: ' . mysql_error());

}



//Select the Database

mysql_select_db($db_name,$db);

$resulting = mysql_query("select * from results", $db); 

//Create an array

$json_response = array();

while ($row = mysql_fetch_array($resulting, MYSQL_ASSOC)) {

    $row_array['home_team_name'] = $row['home_team_name'];

    $row_array['away_team_name'] = $row['away_team_name'];

    $row_array['home_goals_for'] = $row['home_goals_for'];

    $row_array['away_goals_for'] = $row['away_goals_for'];

    //push the values in the array

    array_push($json_response,$row_array);

   $json = json_encode($json_response);


//echo $json;


$json_result = print_r((array) json_decode($json,true));

echo $json_result;

}

留下这个:jay-o.uk/config/config.php

这可能是因为我指的是mySql中的视图而不是表格吗?

答案 2 :(得分:0)

嗯..也许你需要这个代码:

// Create an array
$items = array();

while ($row = mysql_fetch_array($resulting, MYSQL_ASSOC)) {
    /* print new insert */
    var_dump('new item');

    /* add a new item */
    $items[] = array(
        'home_team_name' => $row['home_team_name'],
        'away_team_name' => $row['away_team_name'],
        'home_goals_for' => $row['home_goals_for'],
        'away_goals_for' => $row['away_goals_for']
    );
}

$json_response = json_encode($items);
print $json_response;

如果$ json_response是一个空数组,可能问题是查询没有返回任何行。

答案 3 :(得分:0)

关于一个更简单的解决方案,将数据库中的查询限制为1:

    $results = mysql_query('SELECT * FROM table_name ORDER BY field_name LIMIT 1');