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我加入了 4 个表，3 个表具有相同的列名，称为“备注”。如何显示不同表格的备注结果？这是查询

$qry = mysql_query("SELECT * FROM setup_candidate AS SC,
                                        setup_candidate_degree AS SCD,
                                        setup_candidate_work_experience AS SCWE,
                                        transaction_counselor_candidate_assignment AS TCCA
                                    WHERE SC.Candidate_No = SCD.Candidate_No 
                                    AND SCD.Candidate_No = SCWE.Candidate_No
                                    AND SCWE.Candidate_No = TCCA.Candidate_ID
                                    AND SC.Candidate_No = '$id' 
                                    ");
                    if(!$qry){ echo "problem"}

当我在页面中显示结果时，如何处理特定的备注表？

例如我做了

$row = mysql_fetch_array($qry);

echo $row['Remarks'];

假设我想获得 setup_candidate_degree 表的备注？你能帮助我吗 ？谢谢XD

4

1 回答 1

1

  1. Add alias for those fields.

    SELECT *, setup_candidate_degree.Remaks as setup_candidate_degree_Remaks

  2. Get by number from $row array.

于 2013-11-01T04:14:15.720 回答