从数据库错误中选择mysqli_query / mysqli_fetch_array

时间:2014-02-13 15:00:27

标签: php sql select mysqli

我正试图从数据库中获取此信息:

$query = "SELECT * FROM station_control WHERE station_no > 0 ORDER BY station_ord";

然后说它期望这两行中的参数:

$station_result=  mysqli_query($query);
$num=  mysqli_fetch_array($station_result);

这就是我要输出的内容,基本上是从名为station_ord的数据库中提取每个站名,名称在station_name下:

$i=0;while ($i < $num) {
   $station_name1=mysqli_result($station_result,$i,"station1_name");
   $station2_name=  mysqli_result($station_result, $i,"station2_name");
   $station3_name=  mysqli_result($station_result, $i,"station3_name");
   $station4_name=  mysqli_result($station_result, $i,"station4_name");
   $station5_name=  mysqli_result($station_result, $i,"station5_name");

   echo "<b>
    $station1_name $station2_name2</b>    <br>
    $station3_name<br>
    $station1_name4_name<br>
    $station5_name<hr>    <br>";

   $i++;
}

我还没有放入station_name,因为我很困惑我在哪里以及如何配置它。 任何想法如何帮助?

2 个答案:

答案 0 :(得分:0)

这样的事情应该有效。

$station_result=  mysqli_query($query);
$results=  mysqli_fetch_array($station_result);
foreach ($results as $row) {
  // echo whatever you want here
  echo $row['station_name'];
}

答案 1 :(得分:0)

不确定您尝试做什么,但请尝试

<?php
$station_result=  mysqli_query($query);
$records =  mysqli_fetch_array($station_result);
foreach ($records as $record) :?>
      <b><?php echo $record['station_name']; ?></b><br/>
<?php endforeach; ?>

参考:http://ch1.php.net/manual/en/mysqli.query.phphttp://www.php.net/manual/en/mysqli-result.fetch-array.php示例