我需要使用2个表,我尝试使用内部联接,4个结果的位置,它只显示两个表的结果,我粘贴两个查询,请看它并帮助我,我将感谢所有人。
<?
$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'];
$dest="uploads";
?>
当我打印/ echo
时$es_touring_city = $row5['es_touring_city'];
显示4结果。
但是当我在第二个查询中使用$ es_touring_city时,它只显示带有图像的2个结果。
如果我不是很清楚说话。
谢谢,
答案 0 :(得分:0)
将所有内容加入到单个查询中。对结果进行排序,以便同一+----------------------+
| result |
+----------------------+
| ["ROOT1","N1","SN2"] |
| ["ROOT1","N2","SN3"] |
+----------------------+
的所有行都在一起。然后检查es_id
何时更改,并开始新行。
$row['es_id']