我在我的脚本中有这个代码,它只显示我的第一行..任何解决方案?(这不是我的整个脚本)
$sql = mysql_query ('select * from todo');
if (!$sql) {
die('Invalid query: ' . mysql_error());
}
$taskrow = mysql_query ('select * from todo');
if($_SERVER['REQUEST_METHOD'] == 'POST'){
$taak = $_POST['taak'];
$beschrijving = $_POST['beschrijving'];
$categorie = $_POST['categorie'];
$prioriteit = $_POST['prioriteit'];
$datum = $_POST['datum'];
$query = mysql_query("INSERT INTO todo (taak, beschrijving, categorie, prioriteit, datum) VALUES ('$taak', '$beschrijving', '$categorie', '$prioriteit', '$datum')") or die(mysql_error());
} ?>
<?php $task = mysql_fetch_assoc($taskrow); ?>
<td><?php echo $task['taak']; ?></td>
<td><?php echo $task['beschrijving']; ?></td>
<td><?php echo $task['categorie']; ?></td>
<td><?php echo $task['prioriteit']; ?></td>
<td><?php echo $task['datum']; ?> </td>
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小智 6
通过做:
$task = mysql_fetch_assoc($taskrow);
您将$task变成一个对象,其中包含数据库查询中所选行的数据.
所以,你需要遍历那个对象,以便玩每一行的数据......
所以尝试:
while ($task = mysql_fetch_assoc($taskrow)){?>
<td><?php echo $task['taak'];?></td>
<td><?php echo $task['beschrijving']; ?> </td>
<td><?php echo $task['categorie']; ?></td>
<td><?php echo $task['prioriteit']; ?></td>
<td><?php echo $task['datum']; ?> </td>
<? } ?>
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