Question

我一直试图弄清楚这两天并且没有运气。我正在使用PHP和Javascript，我正在尝试访问另一个页面上的Javascript变量。

这是我的Javascript代码，它只提醒用户什么是taskID ...

<script>
function myFunction(taskID)
{
alert("Task #" + taskID);

}
</script>

这是我的PHP

//select all tasks for this report
            $tasks = mysqli_query($conn,"SELECT taskID, hoursSpent, workDone, plannedTasks FROM Tasks where reportID = $report_id AND projID = $proj_id");
            $task_count = 1;
            while($row1 = mysqli_fetch_array($tasks))
            {
                $taskID = $row1['taskID'];
                //display the task information
                echo "<strong id='tasks'>Task #" . $task_count . "</strong>";
                echo " - <a href = '#' onclick ='myFunction($taskID)'>Edit</a><br>";


                echo "Hours Spent: " . $row1['hoursSpent'] . "<br><br>"; 
                echo "Accomplished: <br>" . $row1['workDone'] . "<br><br>";
                echo "Planned for Next week: <br>" . $row1['plannedTasks'] . "<br>";
                echo "<br>";
                $task_count++;
            }

如何将该taskID传递给另一个页面，以便我可以编辑任务信息？

谢谢！

Answer 1

只需改变一下：

echo " - <a href = '#' onclick ='myFunction($taskID)'>Edit</a><br>";

要

echo " - <a href = 'editTask.php?id=".$taskID."'>Edit</a><br>";

你在那儿。只需确保在editTask.php文件中检查id：

$editID = (int) $_GET['id'];//<-- $editID will hold an int, that is the id of the record the user wanted to edit.

除此之外：很高兴看到你正在使用mysql的最新扩展。不过，代码如下：

mysqli_query($conn,"SELECT taskID, hoursSpent, workDone, plannedTasks FROM Tasks where reportID = $report_id AND projID = $proj_id");

让你容易受到注射攻击。请查看准备好的陈述。

$tasks = mysqli_query($conn, 'SELECT taskID, hoursSpent, workDone, plannedTasks FROM Tasks WHERE reportID = ? AND projID = ?');
mysqli_stmt_bind_param($tasks, 'i', $reportID);//replace i with s if should be a string etc...

一旦您的构建基于客户端数据进行查询（您将使用此方法执行此操作：使用url获取ID ...），这是保护自己免受非常常见威胁的最简单方法。

check this link
This one
当然，this one可以了解注射是什么

Answer 2

喜欢这个

<script>
function myFunction(taskID) {
  location="nextpage.php?task="+taskID;
  return false;

}
</script>



echo " - <a href = '#' onclick ='return myFunction("'.$taskID.'")'>Edit</a>

或更简单

echo " - <a href='nextpage.php?task='.$taskID.'">Edit</a>

将Javascript或PHP变量传递到另一个页面

2 个答案: