我有一个由 MySQL 表填充的表,我正在尝试将该表中的选定行插入到另一个表中。
表中的前三个字段来自其他两个表,具体取决于字段值。
下拉选项由用户使用以下方式提供:
$sql = "SELECT project_proposals.id, project_proposals.title, project_proposals.description, project_proposals.academicname, flux_student_records.studentname, flux_student_records.id, flux_student_records.programme, flux_student_records.academicdiscipline FROM project_proposals JOIN flux_student_records ON project_proposals.academicdiscipline = flux_student_records.academicdiscipline
WHERE project_proposals.academicdiscipline = '$academicdiscipline' AND flux_student_records.studentname = '$studentname'";
$retval = mysql_query($sql) or die(mysql_error());
将这些多个选定行插入到另一个表中的最佳方法是什么?谢谢。
最佳答案
“选择进入”是正确的选择。
SELECT into YourNewTable select project_proposals.id, project_proposals.title, project_proposals.description, project_proposals.academicname, flux_student_records.studentname, flux_student_records.id, flux_student_records.programme, flux_student_records.academicdiscipline FROM project_proposals JOIN flux_student_records ON project_proposals.academicdiscipline = flux_student_records.academicdiscipline
WHERE project_proposals.academicdiscipline = '$academicdiscipline' AND flux_student_records.studentname = '$studentname'";
希望您觉得这很有用。
关于PHP MySQL从下拉框选择中插入多个值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31656369/