Question

嘿伙计们，第一次使用stackoverflow。

你能帮我调试吗？

这是问题所在，这个查询是从我的数据库中选择所有行，由于某种原因，它只输出第二行。

  $top10_query = "SELECT * FROM kicks";
  $result = mysqli_query($cxn, $top10_query) or die("Couldn't execute query.");
  $row = mysqli_fetch_assoc($result);

  $rating = $row['rating'];
  $description = $row['description'];
  $completed = $row['completed'];
  $userid = $row['userid'];
  $posted = $row['posted'];

  while($row = mysqli_fetch_assoc($result)) {
   echo "<tr>";
    echo "<td class='rating'>" . $rating . "</td>";
    echo "<td class='description'>" . $description . " </td>";
    echo "<td class='completed_" . $completed . "'>" . $completed  . "</td>";
    echo "<td class='author'>";
     echo "Posted by: <a href='profile?userid=" . $userid . "'>" . $userid . "</a><br />";
     echo "on "; echo $posted;
    echo "</td>";
   echo "</tr>";
  }

Answer 1

您循环遍历行集，但从不多次检索其值。您从第一行中提取了所有值，并将其缓存在此处：

$rating = $row['rating'];
$description = $row['description'];
$completed = $row['completed'];
$userid = $row['userid'];
$posted = $row['posted'];

将此代码移动到循环中，然后删除第一次获取。

Answer 2

您需要在$rating循环中更新$description，while等：

<?php

  $top10_query = "SELECT * FROM kicks";
  $result = mysqli_query($cxn, $top10_query) or die("Couldn't execute query.");

  while($row = mysqli_fetch_assoc($result)) {
   $rating = $row['rating'];
   $description = $row['description'];
   $completed = $row['completed'];
   $userid = $row['userid'];
   $posted = $row['posted'];

   echo "<tr>";
    echo "<td class='rating'>" . $rating . "</td>";
    echo "<td class='description'>" . $description . " </td>";
    echo "<td class='completed_" . $completed . "'>" . $completed  . "</td>";
    echo "<td class='author'>";
     echo "Posted by: <a href='profile?userid=" . $userid . "'>" . $userid . "</a><br />";
     echo "on "; echo $posted;
    echo "</td>";
   echo "</tr>";
  }

?>

或者，当然，您可以内联$rating等，而不是写$row['rating']。

注意：您可能希望在将变量插入HTML之前通过htmlspecialchars运行变量。否则，像<script>alert('hacked');</script>这样的描述可能会执行一个脚本，让自己受到XSS攻击。

您也可以使用extract。但是，我不建议您这样做，因为它可能会给其他开发人员带来问题和困惑：

<?php

  $top10_query = "SELECT * FROM kicks";
  $result = mysqli_query($cxn, $top10_query) or die("Couldn't execute query.");

  while($row = mysqli_fetch_assoc($result)) {
   extract($row);

   echo "<tr>";
    echo "<td class='rating'>" . $rating . "</td>";
    echo "<td class='description'>" . $description . " </td>";
    echo "<td class='completed_" . $completed . "'>" . $completed  . "</td>";
    echo "<td class='author'>";
     echo "Posted by: <a href='profile?userid=" . $userid . "'>" . $userid . "</a><br />";
     echo "on "; echo $posted;
    echo "</td>";
   echo "</tr>";
  }

?>

Answer 3

变量$rating等没有“绑定”到表达式$row['rating']等。一旦设置，它们将永远采用这些值，除非您再次修改它们。

有关详细信息，请参阅PHP: Assignment Operators。

尝试将它们重写为：

  $top10_query = "SELECT * FROM kicks";
  $result = mysqli_query($cxn, $top10_query) or die("Couldn't execute query.");

  while($row = mysqli_fetch_assoc($result)) {
    $rating = $row['rating'];    // <-- use the new value every time a row is fetched.
    $description = $row['description'];
    $completed = $row['completed'];
    $userid = $row['userid'];
    $posted = $row['posted'];

   echo "<tr>";
    echo "<td class='rating'>" . $rating . "</td>";
    echo "<td class='description'>" . $description . " </td>";
    echo "<td class='completed_" . $completed . "'>" . $completed  . "</td>";
    echo "<td class='author'>";
     echo "Posted by: <a href='profile?userid=" . $userid . "'>" . $userid . "</a><br />";
     echo "on "; echo $posted;
    echo "</td>";
   echo "</tr>";
  }

MySQL没有拉动所有行

3 个答案: