php - php中Mysql内连接搜索

Question

这是我的选择查询

 $raw_results = mysql_query("SELECT *,student_data.student_id,student_data.image,student_data.firstname, student_data.middleI,
            student_data.lastname, 
            student_data.address, student_data.birthday,
            section.section_name, department.dept_name, year_graduated.year, course.course_name,
            year_graduated.publish_id  FROM student_data        
    INNER JOIN course ON student_data.course_id=course.course_id
    INNER JOIN section ON student_data.section_id=section.section_id
    INNER JOIN department ON student_data.dept_id=department.dept_id
    INNER JOIN year_graduated ON student_data.year_id=year_graduated.year_id WHERE (`firstname` LIKE '%".$query."%') OR (`lastname` LIKE '%".$query."%') OR (`birthday` LIKE '%".$query."%') OR (`section_name` LIKE '%".$query."%') OR (`course_name` LIKE '%".$query."%') OR (`year` LIKE '%".$query."%') OR (`address` LIKE '%".$query."%') AND publish_id=2")  ;

我只想显示year_graduated.publish_id等于2的student_data。当我运行上面的代码时，它不会过滤学生的publish_id。

Answer 1

我认为括号有错误。 将开头放在 (firstname LIKE '%".$query."%') 之前，将结束放在 AND ANDpublish_id=2

之前

就这样

$raw_results = mysql_query("SELECT *,student_data.student_id,student_data.image,student_data.firstname, student_data.middleI,
            student_data.lastname, 
            student_data.address, student_data.birthday,
            section.section_name, department.dept_name, year_graduated.year, course.course_name,
            year_graduated.publish_id  FROM student_data        
    INNER JOIN course ON student_data.course_id=course.course_id
    INNER JOIN section ON student_data.section_id=section.section_id
    INNER JOIN department ON student_data.dept_id=department.dept_id
    INNER JOIN year_graduated ON student_data.year_id=year_graduated.year_id WHERE ((`firstname` LIKE '%".$query."%') OR (`lastname` LIKE '%".$query."%') OR (`birthday` LIKE '%".$query."%') OR (`section_name` LIKE '%".$query."%') OR (`course_name` LIKE '%".$query."%') OR (`year` LIKE '%".$query."%') OR (`address` LIKE '%".$query."%')) AND publish_id=2")