php嵌套while循环。内循环只能工作一次

Question

这是代码。任何帮助？

<?php 

    //connect to the database 
     $con=mysql_connect("localhost","root","") or die("Error in connection"); 
     mysql_select_db("riffadb",$con)or die("Error in database"); 
    // 
     $sql1 ="select category_id,category_code from category"; 
     $res1 = mysql_query($sql1); 

     $sql2 = "select category_id1,product_id from product"; 
     $res2 = mysql_query($sql2); 
     $i=0; 
     $j=1000; 
     while($row1 = mysql_fetch_array($res1)) 
     { 
      echo $row1['category_code']; 
      echo '<br/>'; 
      while($row2 = mysql_fetch_array($res2)) 
      { 
       if($row1['category_code'] == $row2['category_id1']) 
       { 
        $sql3 = "update product set category_id = ".$row1['category_id']." where product_id = ".$row2['product_id']; 
        echo '<br/>'; 
        mysql_query($sql3); 
       } 
      } 
      $i++; 
     } 
     echo $i; 
    ?> 

Answer 1

我认为你需要重新查询RES2外环内，或至少复位指针。第二次通过外部循环，你在res2中没有剩下任何东西，因为你已经到了最后

另外，如果你在外部循环中设置res2查询并且添加一个where子句，以便您只与res1中当前行的类别ID匹配。

希望这会有所帮助。

Answer 2

广场这里面第一个循环：

$sql2 = "select category_id1,product_id from product"; 
$res2 = mysql_query($sql2); 

Answer 3

试试这个：

<?php 

//connect to the database 
    $con=mysql_connect("localhost","root","") or die("Error in connection"); 
    mysql_select_db("riffadb",$con)or die("Error in database"); 
// 
    $sql1 ="select category_id,category_code from category"; 
    $res1 = mysql_query($sql1); 


    $i=0; 
    $j=1000; 
    while($row1 = mysql_fetch_array($res1)) 
    { 
     echo $row1['category_code']; 
     echo '<br/>'; 
      // execute second query under the first loop 
     $sql2 = "select category_id1,product_id from product"; 
     $res2 = mysql_query($sql2); 

     while($row2 = mysql_fetch_array($res2)) 
     { 
      if($row1['category_code'] == $row2['category_id1']) 
      { 
       $sql3 = "update product set category_id = ".$row1['category_id']." where product_id = ".$row2['product_id']; 
       echo '<br/>'; 
       mysql_query($sql3); 
      } 
     } 
     $i++; 
    } 
    echo $i; 
?>