PHP 如何创建所有任务的列表

发布于 2024-11-30 10:23:21 字数 366 浏览 1 评论 0原文

好吧，我有一个任务表，该表有一个外键，即“projectID”，

我正在选择该表中具有相同 projectID 的所有行。但我现在想将结果输出到列表中 ("

//Select tasks
$sql = "SELECT * FROM tasks WHERE projectID = '".$project_ID."'";
$result5 = $db->sql_query($sql);
$data5 = mysql_fetch_assoc($result5);

在此输入图像描述

原文

Alright, I have a table of tasks this table has a foreign key which is the "projectID"

I am selecting all the rows within that table that have the same projectID. But I now want to output the results in a list ("<li></li>")

//Select tasks
$sql = "SELECT * FROM tasks WHERE projectID = '".$project_ID."'";
$result5 = $db->sql_query($sql);
$data5 = mysql_fetch_assoc($result5);

enter image description here

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水溶 2024-12-07 10:23:21

您将需要迭代每一行并输出 taskList 列值：

$sql = "SELECT * FROM tasks WHERE projectID = '".$project_ID."'";
$result5 = $db->sql_query($sql);

$data5 = mysql_fetch_assoc($result5);
// this will let you handle an empty result set without a count.
if( $data5 )
{
    // opening the list only if there are things to put there
    echo "<ul >";
    do
    {
        // output the value from one row.
        echo "<li>" . $data5['taskName'].'</li>';
    }while($data5 = mysql_fetch_assoc($result5));
    echo "</ul>";
}
else
{ 
    echo 'No tasks found!';
}

You're going to need to iterate through each of the rows and output the taskList column value:

$sql = "SELECT * FROM tasks WHERE projectID = '".$project_ID."'";
$result5 = $db->sql_query($sql);

$data5 = mysql_fetch_assoc($result5);
// this will let you handle an empty result set without a count.
if( $data5 )
{
    // opening the list only if there are things to put there
    echo "<ul >";
    do
    {
        // output the value from one row.
        echo "<li>" . $data5['taskName'].'</li>';
    }while($data5 = mysql_fetch_assoc($result5));
    echo "</ul>";
}
else
{ 
    echo 'No tasks found!';
}

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