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我遇到了一个有趣的问题，即使用一个带有几个连接的选择查询从 mysql 表中检索数据。

1）查询：

$task_details = "SELECT tasks.task, ";
$task_details = $task_details . "tasks.description,";
$task_details = $task_details . "tasks.finishby, ";
$task_details = $task_details . "responsibles.full_name, ";
$task_details = $task_details . "task_assignments.completed, ";
$task_details = $task_details . "tasks.id, ";
$task_details = $task_details . "responsibles.user_id ";
$task_details = $task_details . "FROM tasks,task_assignments,responsibles ";
$task_details = $task_details . "WHERE ";
$task_details = $task_details . "tasks.id = task_assignments.id_task AND ";
$task_details = $task_details . "responsibles.id = task_assignments.id_assignee AND ";
$task_details = $task_details . "tasks.id = $id_task;";

$task_details_q = mysql_query($task_details) or die(mysql_error());

1a）结果示例查询： 

SELECT tasks.task, tasks.description, tasks.finishby, responsibles.full_name, task_assignments.completed, tasks.id, responsibles.user_id
FROM tasks, task_assignments, responsibles
WHERE tasks.id = task_assignments.id_task
AND responsibles.id = task_assignments.id_assignee
AND tasks.id =19

2）HTML/PHP代码：

<table class="task_table">
        <thead>
            <th>Task</th>
            <th>Description</th>
            <th>Due date</th>
            <th>Person</th>
            <th>Completed</th>
        </thead>
        <?php

        $even = false;
        $trow = "";
while($row = mysql_fetch_array($task_details_q))
        {

            $trow = $trow . "<tr";
             if($even) $trow = $trow . " style=\"background-color: #f2f2ed; \"";
            $trow = $trow. ">";
            $trow = $trow . "<td >$row[0]</td>";
            $trow = $trow . "<td>" . $row[1] . "</td>";
            $trow = $trow . "<td>" . date('d-m-Y',$row[2]) . "</td>";
            $trow = $trow . "<td>$row[3]</td>";
            $trow = $trow . "<td style=\"text-align: center;\" >";
            if($row[4] > 0)
            {
                $trow = $trow . "<a href=\"javascript:modifyCompleted('remove','$row[6]',$row[5])\" title=\"Click to change completion of this task by this person\">yes</a>";
            }
            else
            {
                $trow = $trow . "<a href=\"javascript:modifyCompleted('add','$row[6]',$row[5])\" title=\"Click to change completion of this task by this person\">no</a>";
            }
            $trow = $trow . "</td>";
            $trow = $trow . "</tr>";

            $even =! $even;
            $number = $number + 1;
        }
            $trow = $trow . "<tr style=\"border-top: 1px solid #666666;\"><td></td><td></td><td></td><td></td>";
            $trow = $trow . "<td>";
            $trow = $trow . "<a href=\"javascript:modifyCompleted('add_all','all',$task_details_array[5])\" title=\"Click to complete all\">Complete all</a>";
            echo $trow;
        ?>
    </table><br />
    <span style="text-align: center;display:block;font-size: 12px;"><a href="tasks.php">Go back to task overview</a></span>

3）问题/问题：由于某种原因，显示的表格总是省略​​一条记录。我在许多 PHP 脚本中使用了相同（或非常相似的概念），但从未遇到过相同的问题。我认为查询本身不是问题——当我直接对数据库运行它时，它会返回正确数量的值......（我认为）。

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4 回答 4

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2件事：

1）您不应该在每次添加信息时提交答案。您应该单击原始问题下的“编辑”并将新信息添加到问题中。

2）我想如果我澄清如何mysql_query和mysql_fetch_array工作你会看到发生了什么。

当您mysql_query使用“SELECT”调用时，查询它会返回一个resource. 这resource只是对记录集的引用。然后，当您调用它时mysql_fetch_array，resource它将从集合中返回当前记录，并推进记录指针。

所以，当你第一次调用时mysql_query，记录指针指向第一个结果。然后你调用mysql_fetch_array第一条记录作为数组返回，并且指针前进到第二条记录。下次调用mysql_fetch_array此第二条记录时将返回，然后指针将指向第三条记录。

如果没有第 3 条记录，下次调用mysql_fetch_array将找不到对应的记录，会返回 false。

这就是您使用while($row = mysql_fetch_array($task_details_q,MYSQL_NUM)). 您将结果放入变量中$row并推进结果指针，然后使用$row. 最终，您会将指针前进到最后一个结果之外，并且$row将为 false，这将阻止您的 while 循环前进。

现在我已经了解了理论，这里是您的代码发生了什么（我将删除不相关的代码//...并在此过程中添加我自己的注释）：

$task_details_q = mysql_query($task_details) or die(mysql_error());
//now you have a resource $task_details_q, it points to the first result


$task_details_array = mysql_fetch_array($task_details_q,MYSQL_NUM);
//you retreive the first result, and advance the pointer to the second result

for($x=0;$x < sizeof($task_details_array);$x++)
{
            //you perform operations (echo'ing in this case) on your first result
    echo $x . ". : " . $task_details_array[$x] . "<br />";
}

//... HTML CODE SKIPPED

$even = false;
$trow = "";

//the first time this while statement is called you place the data from the second result in $row,
        //and advance the pointer to the third result
//the next time you go through the loop you try to place the data from the third result in $row,
        //since you say there are only 2 results to your query $row is simply false. 
        //This causes the while to stop executing and the code to continue on
while($row = mysql_fetch_array($task_details_q,MYSQL_NUM))
{
    //... PRINT TABLE CELLS FROM $row SKIPPED
}

//... REMAINING HTML SKIPPED

我不确定您是否真的需要代码块：

$task_details_array = mysql_fetch_array($task_details_q,MYSQL_NUM);
for($x=0;$x < sizeof($task_details_array);$x++)
{
    echo $x . ". : " . $task_details_array[$x] . "<br />";
}

或者如果您只是添加它进行调试。如果它只是用于调试，请将其删除，您的第一个结果将显示在 while 循环中。如果您需要for执行该循环，请对此答案发表评论，我将编辑我的答案以在forandwhile循环中使用第一个结果。

于 2012-04-19T15:52:28.257 回答
0

由于 php 似乎是正确的，因此问题的原因可能在于 html。您的标题行没有<tr>标签，因此浏览器可能会因为您缺少第一条记录而窒息。

修复此问题后，我建议在 html 验证器中检查您的 html，以确保那里没有更多错误。

于 2012-04-18T21:01:42.207 回答
0

@Ben：这是和之间的完整mysql_query($task_details)代码while($row = mysql_fetch_array($task_details_q))：

 $task_details_q = mysql_query($task_details) or die(mysql_error());
 $task_details_array = mysql_fetch_array($task_details_q,MYSQL_NUM);

for($x=0;$x < sizeof($task_details_array);$x++)
{
    echo $x . ". : " . $task_details_array[$x] . "<br />";
}

?>

<h3>Task details - "<?php echo strtoupper($task_name);  ?>"</h3>
<span class="notifierOK">Table below lists all people assigned to the task - including the status (complete / incomplete). To change person's status click on the 'yes' or 'no' link. If you then go back
to (<a href="tasks.php">previous page</a>) the completion percentage value will be re-calculated.</span><br />

<form id="task_details" method="post" name="task_details" style="margin-left: auto; margin-right: auto;width: 800px;box-shadow:10px 10px 5px #888888;">
<table class="task_table">
        <thead>
            <tr>
            <th>Task</th>
            <th>Description</th>
            <th>Due date</th>
            <th>Person</th>
            <th>Completed</th>
            <tr />
        </thead>
        <?php

        $even = false;
        $trow = "";

        while($row = mysql_fetch_array($task_details_q,MYSQL_NUM))

请注意，我已根据您的建议将“MYSQL_NUM”添加为数组类型，因此它不一定属于那里。

于 2012-04-19T10:15:50.993 回答
0

我解决了，我有同样的问题。

$run_query = mysqli_query($conn, $stores);

                if ($run_query === false){
                    //error
                }else if (mysqli_num_rows($run_query)){
                    $row = mysqli_fetch_array($run_query);
                    echo 'bla bla bla' //on this echo I needed to show the data once.
                    $run_query =  null;
                    $run_query = mysqli_query($conn, $stores);
                    while ($row = mysqli_fetch_array($run_query)) { //and here I needed to show up the loops of the results of my query.}

所以...我只是重置了运行查询的变量并重新运行了搜索。它对我有用！;)

于 2017-03-16T13:50:16.787 回答