Question

我想使用onclick Id显示两个表中的元素。这些是我的表格：

Project_tech：

project_tech_id  project_tech_name
        1            Mechanical
        2            Civil
        3            Computers

project_detail：

id      title       content       project_tech_id
 1     sample        abcd              3
 2     sample2       efgh              1
 3     sample3       xyz               3

使用project_tech_id在两个表之间存在关系。使用表单将值插入到project_detail中。现在我想显示标题，内容，project_tech_name其中project_tect_id = 3来自project_detai表意味着我想要两个显示两列数据... 如何使用一个查询显示它实际上我使用这个查询，但我收到错误：

$prj_id = $_GET['id'];

$sql = mysql_query("SELECT pt.project_tech_name,
    pd.title, 
    pd.content,
    pd.video_url,
    FROM
      project_details pd,
      project_tech pt
      WHERE
      pt.project_tech_Id = pd.project_tech_Id AND
      pt.project_tech_Id in" .$prj_id);

请帮帮我......

Answer 1

您应该使用=而不是

Answer 2

尝试执行它

mysql_query("SELECT pt.project_tech_name,pd.title,pd.content,pd.video_url
    FROM
      project_details pd Left Join project_tech pt
      ON
      pt.project_tech_Id = pd.project_tech_Id AND
      pt.project_tech_Id = " .$prj_id);

Answer 3

使用 JOIN ，您可以这样做：

$prj_id = $_GET['id'];

$sql = mysql_query("SELECT pt.project_tech_name,
            pd.title, pd.content, pd.video_url
            FROM project_details pd JOIN project_tech pt
            ON pd.project_tech_id = pt.project_tech_id
            WHERE pt.project_tech_Id = {$prj_id}");

Answer 4

$sql = mysql_query("SELECT pt.project_tech_name,
    pd.title, 
    pd.content,
    pd.video_url
    FROM
      project_details pd,
      project_tech pt
      WHERE
      pt.project_tech_Id = pd.project_tech_Id 
      AND
      pt.project_tech_Id in (" .$prj_id.")");

或

pt.project_tech_Id = " .$prj_id);

//我刚刚删除了From子句之前的逗号

Answer 5

你的查询可以是：这是一个优化的，因为它使用LEFT JOIN。

$sql = mysql_query("SELECT pt.project_tech_name,
    pd.title, 
    pd.content,
    pd.video_url,
    FROM project_details pd
     LEFT JOIN project_tech pt 
      ON pt.project_tech_Id = pd.project_tech_Id
    WHERE  pt.project_tech_Id = {$prj_id} ");

Answer 6

在您的SQL查询中，有一个字段名称“pd.video_url”，它不在您的表结构中，并且在此字段值之后还有一个额外的逗号（，）。

第二个问题是，使用“=”而不是“in”

如果有多个“project_tech_Id”，则可以使用“in”之类的



$prj_id = (1,4,7,2);

$sql = mysql_query("SELECT pt.project_tech_name,
    pd.title, 
    pd.content,
    pd.video_url,
    FROM
      project_details pd,
      project_tech pt
      WHERE
      pt.project_tech_Id = pd.project_tech_Id AND
      pt.project_tech_Id in(" .$prj_id .")".);

$prj_id = (1,4,7,2); $sql = mysql_query("SELECT pt.project_tech_name, pd.title, pd.content, pd.video_url, FROM project_details pd, project_tech pt WHERE pt.project_tech_Id = pd.project_tech_Id AND pt.project_tech_Id in(" .$prj_id .")".);

Answer 7

试试这个：

"SELECT 
project_tech.project_tech_name,
project_detail.title, 
project_detail.content,
project_detail.video_url

FROM project_tech
JOIN project_details
ON 
project_tech.project_tech_Id = project_details.project_tech_Id 
AND
project_tech.project_tech_Id = $prj_id"

使用id从两个表中获取数据

7 个答案: