Question

我有一个非常不寻常的排序问题。我必须按照我现在在数据库中预先定义的位置来安排项目：

sid - auto increment
name - just something to distinguish varchar 100
sorting - this is where we define the exact position of the item

如果我们向任何产品提供sorting = 2，无论如何，它都将处于第二位。但它应该按照排序的定义。

我无法使用order by sorting desc排列项目，因为大多数项目都没有分配任何位置。由于某些已分配值的条目具有sorting = 0，因此根据计划，所有排序值不等于零的项目应放在其他项目之前。

这是当前的代码：

 <?php 
  $conn=mysqli_connect("localhost","root","","work");
   $getthedetails=mysqli_query($conn,"select sid,sorting,name from sorting order by sorting asc");
  while($thelistis=mysqli_fetch_array($getthedetails)){ ?>
  <div style="width:100px;font-weight:bold;color:#000;font-size:19px;font-   family:arial;height:100px;padding:49px;background:silver;float:left;margin:10px;">
    <?=$thelistis['sorting']?>
   </div>
   <?php } ?>

理想情况下，我希望排序为1,0,0,4,5,0,0,8。

我想要的是this。

Answer 1

请尝试以下

$getthedetails=mysqli_query($conn,"select id,sorting,name from tbl_sorting order by sorting asc");
$arrayList = array();
while($thelistis=mysqli_fetch_array($getthedetails)){
    $arrayList[] = $thelistis;
}

foreach($arrayList as $key=>$list){
    if($list['sorting'] != 0){
        //Keep current element of array
        $temp = $arrayList[$key];
        //Make current element same as with the one to swap
        $arrayList[$key] = $arrayList[$list['sorting']-1];
        //Swap the kept one to its place
        $arrayList[$list['sorting']-1] = $temp;
    }
}


foreach($arrayList as $key=>$list){?>

<div style="width:100px;font-weight:bold;color:#000;font-size:19px;font-   family:arial;height:100px;padding:49px;background:silver;float:left;margin:10px;">
    <?php echo $list['sorting']?>
</div>
<?php } ?>

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不寻常的数据排序 - 按固定数据ID

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