Question

我是新来的，如果有人能帮我解决这个问题，我会很高兴。我想从数组$ data中显示不在数组$ time中的时间选择。它应该显示12个中的11个，因为这些数组之间有一个相同的项目。但是它只显示其中一个。希望有人可以帮助我。先谢谢你！

<?php

$q = "SELECT masa FROM pra_temujanji where status = 'Diterima' and hari = '$hari' and Doktor = '$Doktor'";
$select = mysql_query($q);
$time = mysql_fetch_array($select);

    $qry = "SELECT Masa FROM masa_temujanji";
    $qid = mysql_query ($qry);
    $data = mysql_fetch_array($qid);

    foreach($time as $elem) $time[$elem['masa']];
    foreach($data as $elem) $data[$elem['Masa']];
    ksort($data); ksort($time);

    $waktu = array_diff($data, $time);
    ?>

    <option value="<?php print_r($waktu['Masa'])?>"><?php print_r($waktu['Masa'])?></option>


    </select>

我真的不知道上一篇文章是如何替换的，所以我只想在这里编辑。我现在可以显示数组，但输出是这样的。

9时00分零零秒 09:30:00 。。。注意：身份不明的索引：Masa ...... 10:00:00 12:00:00

我不希望出现错误。它实际上是两个数组之间的array_diff。怎么可以跳过？

代码：

<?php   
    $q = "SELECT masa FROM pra_temujanji where status = 'Diterima' and hari = '$hari' and Doktor = '$Doktor'";
    $select = mysql_query($q);
    while($time = mysql_fetch_array($select)):

    $qry = "SELECT Masa FROM masa_temujanji";
    $qid = mysql_query ($qry);
    while ($data = mysql_fetch_array($qid)):

    $waktu = array_diff($data, $time);
    ?>

    <option value="<?php print_r($waktu['Masa'])?>"><?php print_r($waktu['Masa'])?></option>

    <?php
    endwhile;
    ?>   
    </select>
    <?php
    endwhile;
    ?>

感谢您的帮助。这是解决我的问题的代码

<?php  

$q = "SELECT masa FROM pra_temujanji where status = 'Disahkan Lulus' and hari = '$hari' and Doktor = '$Doktor'";
$select = mysql_query($q);
while ($time = mysql_fetch_array($select)){
    $masa[] = $time['masa'];
}
if(isset($masa)){

$qry = "SELECT Masa FROM masa_temujanji";
$qid = mysql_query ($qry);
while ($data = mysql_fetch_array($qid)){
$Masa[] = $data['Masa'];

}
$waktu = array_diff($Masa, $masa);
foreach( $waktu as $key => $value){
?>

<option value="<?php  echo $value ?>"><?php  echo $value ?></option>

  <?php } }
  else{
      $qry = "SELECT Masa FROM masa_temujanji";
$qid = mysql_query ($qry);
while ($data = mysql_fetch_array($qid)){
$Masa[] = $data['Masa'];

}
//$waktu = array_diff($Masa, $masa);
foreach( $Masa as $key => $value){
?>

<option value="<?php  echo $value ?>"><?php  echo $value ?></option>
<?php

}}
  ?>

再次感谢你。

array_diff只打印给出一个结果而不是很多

0 个答案: