Question

好的，我有一个简单的php select脚本，我有下面的代码来定义变量。

$result = mysqli_query($con,"SELECT * FROM practice_sheets WHERE student_name='$_SESSION[SESS_FIRST_NAME] $_SESSION[SESS_LAST_NAME]'");

   $numrows = mysqli_num_rows($result);
   $id = $row['id'];
   $total_min = $row['total_min'];
   $due_date = $row['due_date'];

然后我有：

echo "<td> <a href='account/practiceSheets?id='$id'> <i class='icon-eye-open'> </i> </a> </td>";

这应该是在点击<a>时将php select脚本中的变量传递给url。

我最终得到的是account/practiceSheets?id=没有实际ID。我确信这是一个非常简单的事情，我很抱歉，因为我是PHP的新手，也不知道该怎么称呼它才能在搜索引擎中获得可用的结果！如果它有帮助，我的完整代码如下。

<?php
$con = mysqli_connect("50.63.106.47", "usd309bands", "MacBook1!", "usd309bands");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con, "SELECT * FROM practice_sheets

 WHERE student_name='$_SESSION[SESS_FIRST_NAME] $_SESSION[SESS_LAST_NAME]'");

 $numrows   = mysqli_num_rows($result);
 $id        = $row['id'];
 $total_min = $row['total_min'];
 $due_date  = $row['due_date'];

 if ($numrows == 0) {
   echo "<div class='alert alert-danger'>";
   echo "No Entries, See your instructor for details.";
   echo "</div>";
 } else {

   echo "<table class='mws-table table-striped table-hover'>";
   echo "<thead align='center'>";
   echo "<tr>";
   echo "<th>Sheet Number</th>";
   echo "<th>Total Minutes</th>";
   echo "<th>Due Date</th>";
   echo "<th>View</th>";
   echo "</tr>";
   echo "</thead>";
   echo "<tbody align='center'>";

   while ($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>" . $row['id'] . "</td>";
    echo "<td>" . $row['total_min'] . "</td>";
    echo "<td>" . $row['due_date'] . "</td>";
    echo "<td> <a href='account/practiceSheets?id='$id'> <i class='icon-eye-open'> </i>  </a> </td>";
    echo "</tr>";
    }
     echo "</tbody>";
     echo "</table>";

    mysqli_close($con);
    }
    ?>

Answer 1

$result = mysqli_query($con,"SELECT * FROM practice_sheets
                                                    WHERE student_name='$_SESSION[SESS_FIRST_NAME] $_SESSION[SESS_LAST_NAME]'");

$numrows = mysqli_num_rows($result);
$id = $row['id'];

看不到$row的来源。

我认为您忘记将MySQL响应提取到关联的数组中。（您可以使用mysqli_fetch_assoc()来完成那个）

这是一个简单的例子：

$result = mysqli_query($con,"SELECT * FROM practice_sheets WHERE student_name='$_SESSION[SESS_FIRST_NAME] $_SESSION[SESS_LAST_NAME]'");
$allRows = mysqli_fetch_assoc($result);
foreach($allRows as $row) {
  echo $row['id'].'<br/>';
}

这个应列出数据库请求中的所有ID。

Answer 2

这个SQL查询可能不会返回任何内容，除非名称存储在您的表中，如此

firstlast

"SELECT * FROM practice_sheets WHERE student_name='$_SESSION[SESS_FIRST_NAME] $_SESSION[SESS_LAST_NAME]'"

你可能想要这样的东西

$_SESSION[SESS_FIRST_NAME] . " " . $_SESSION[SESS_LAST_NAME]

返回类似

的名称

最后一次

Answer 3

我改变了

echo "<td> <a href='account/practiceSheets?id='$id'> <i class='icon-eye-open'></i></a></td>"

到此：

echo "<td> <a href='account/practiceSheets?id=" . $row["id"] . "'> <i class='icon-eye-open'> </i> </a> </td>";

并获得了预期的结果！

将PHP select元素传递给url

3 个答案: