PHP循環查詢不起作用

Question

我需要用2個表格工作，我嘗試了內部連接，4個結果的位置，它只是顯示了兩個表格的結果，我正在粘貼這兩個查詢，請看它並幫助我，我會感謝你們所有人。

<? 
    $query5 = "SELECT * FROM escorts_touring order by es_tou_id"; 
    $result5 = mysql_query($query5); 
    while($row5 = @mysql_fetch_array ($result5, MYSQL_ASSOC)) 
    { 
     $es_touring_city = $row5['es_touring_city']; 
    } 

?> 

<?php 

     echo $sql="SELECT e.es_id, e.es_sex, e.service_type, 
          e.working_name, t.es_tou_id, t.es_id, 
          t.es_touring_city, t.es_touring_start_date, 
          t.es_touring_end_date 
        FROM escorts AS e 
         INNER JOIN escorts_touring AS t 
          ON e.es_id = t.es_id 
        where es_touring_city = '$es_touring_city'"; 
     $result=mysql_query($sql); 
     $rowcount=mysql_num_rows($result); 
     $counter=0; 
     $count=0; 

     while($row = @mysql_fetch_array ($result, MYSQL_ASSOC)) 
     { 

      if($counter++%4==0)print"</div><div class=\"row\"></div>"; 

      $es_sex =$row['es_sex']; 
      $service_type=$row['service_type']; 
      $working_name=$row['working_name']; 
      $es_id=$row['es_id']; 
      $es_tou_id = $row['es_tou_id']; 
      $es_touring_city = $row['es_touring_city']; 
      $es_touring_start_date = $row['es_touring_start_date']; 
      $es_touring_end_date =$row['es_touring_end_date']; 

      $newstartDate = date("dS F, Y", strtotime($es_touring_start_date)); 
      $newendDate = date("dS F, Y", strtotime($es_touring_end_date)); 

      $query = "SELECT * FROM escorts_image where es_id = $es_id"; 
      $result_image = @mysql_query ($query); 
      $row_image = @mysql_fetch_array ($result_image, MYSQL_ASSOC); 
      $image = $row_image['image']; 
      $dest="uploads"; 
     ?> 

當我打印/回波

$es_touring_city = $row5['es_touring_city']; 

其示出圖4的結果。

但是當我在第二個查詢中使用$ es_touring_city時，它只顯示2個結果與圖像。

如果我不是很清楚說話。

謝謝，

Answer 1

將所有內容加入到單個查詢中。對結果進行排序，以便同一個es_id的所有行都在一起。然後檢查$row['es_id']何時更改，並開始新行。

<?php 

$sql="SELECT e.es_id, e.es_sex, e.service_type, 
      e.working_name, t.es_tou_id, t.es_id, 
      t.es_touring_city, t.es_touring_start_date, 
      t.es_touring_end_date, i.image 
     FROM escorts AS e 
     INNER JOIN escorts_touring AS t ON e.es_id = t.es_id 
     INNER JOIN (SELECT es_id, MAX(image) AS image 
        FROM escorts_image 
        GROUP BY es_id) AS i ON e.id = i.es_id 
     ORDER BY e.es_id"; 
$result = mysql_query($sql); 
$last_esid = null; 
$counter = 0; 

$dest = "uploads"; 

while ($row = mysql_fetch_assoc($result)) { 
    $es_id=$row['es_id']; 
    if ($counter++ == 4 || $es_id != $last_esid) { 
     if ($last_esid) { 
      echo "</div>"; 
     } 
     echo "<div class='row'></div>"; 
     $last_esid = $es_id; 
     $counter = 0; 
    } 

    $es_sex =$row['es_sex']; 
    $service_type=$row['service_type']; 
    $working_name=$row['working_name']; 
    $es_tou_id = $row['es_tou_id']; 
    $es_touring_city = $row['es_touring_city']; 
    $es_touring_start_date = $row['es_touring_start_date']; 
    $es_touring_end_date =$row['es_touring_end_date']; 
    $image = $row['image']; 

    $newstartDate = date("dS F, Y", strtotime($es_touring_start_date)); 
    $newendDate = date("dS F, Y", strtotime($es_touring_end_date)); 

    // print the row 
}