[英]Why does the PrepareStatement not generate correct SQL query?
以下是我的PrepareStatement代碼。 它不會生成正確的SQL查詢。 它也沒有超越1st println-statement。 另外它在查詢中說**未指明**(請參見下文)。
我們怎么解決這個問題呢?
ps1 = con.prepareStatement(
"select stuId, name, relationsName, houseAddress, houseNumber from temp where "
+ " stuId like '?%' and "
+ " sex = '?' and "
+ " name like '?%' and "
+ " age BETWEEN ? and ? and "
+ " relationsName like '?%' "
+ " order by name asc limit 0, 150000 "
);
System.out.println("ps1 Before : " + ps1);
輸出:
ps1之前:com.mysql.jdbc.JDBC4PreparedStatement@14d55de:從temp中選擇stuId,name,relationsName,houseAddress,houseNumber,其中stuId喜歡'?%'和sex ='?' 並命名為'?%'和年齡BETWEEN **未指明**和**未指明**和relationsName喜歡'?%'按名稱排序asc limit 0,150000
它沒有超出這一點。 此外,它說不在查詢中指定 (請參閱為末)。
有什么見解嗎?
ps1.setString(1, stuId);
ps1.setString(2, gender);
ps1.setString(3, name);
ps1.setInt(4, startAge);
ps1.setInt(5, endAge);
ps1.setString(6, relationsName);
System.out.println("ps1 After : " + ps1);
rs = ps1.executeQuery();
因為占位符用單引號括起來,因此使它成為一個值而不再是一個參數。 你應該擺脫它,例如
ps1 = con.prepareStatement(
"select stuId, name, relationsName, houseAddress, houseNumber from temp where "
+ " stuId like ? and "
+ " sex = ? and "
+ " name like ? and "
+ " age BETWEEN ? and ? and "
+ " relationsName like ? "
+ " order by name asc limit 0, 150000 "
);
對於LIKE
語句,你應該在java中連接值,而不是在sql中,
ps1.setString(1, stuId + '%');
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