dongzhanjuan5141 2013-12-16 09:01
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在表格中显示记录

I am having some trouble with my first PHP project, I am trying to get the data from MySQL database (Has 3 records) and display it in tables. Problem is it only seem to display records 2 and 3, it skips the 1st record. Please see my code and display below.

if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM unitstats");

while($row = mysqli_fetch_array($result)) {
  echo "<table border='1' style='color:white'>
  <tr>
  <th>ID</th>
  <th>Name</th>
  </tr>";

  while($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>" . $row['id'] . "</td>";
    echo "<td>" . $row['name'] . "</td>";
    echo "</tr>";
  }
  echo "</table>";
}

enter image description here

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3条回答 默认 最新

  • dsmlf1207915 2013-12-16 09:06
    关注

    you are using two while loop which is unnecessary use following code

    if (mysqli_connect_errno())
          {
          echo "Failed to connect to MySQL: " . mysqli_connect_error();
          }
    
          echo "</table>";
    
    
    echo "<table border='1' style='color:white'>
    <tr>
    <th>ID</th>
    <th>Name</th>
    </tr>";
    $result = mysqli_query($con,"SELECT * FROM unitstats");
    
    while($row = mysqli_fetch_array($result))
      {
    
    
    echo "<tr>";
      echo "<td>" . $row['id'] . "</td>";
      echo "<td>" . $row['name'] . "</td>";
      echo "</tr>";
    }
      echo "</table>";
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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