Jer*_*unx 11 java mysql database jdbc
你好,我正试图在没有匹配的学生时犯错......它会显示如下这个
No matching records found,我希望列名仍然相同,但仍然没有把它搞清楚......有人可以告诉我这是不是对??
继承我的功能......我在那里添加评论我把错误...但我不知道如何获得列名
public void SearchTableStudent() {
String tempSearchValue = searchStudent.getText().trim();
boolean empty = true;
sql = "SELECT student_id as 'Student ID',"
+ "concat(lastname, ' , ', firstname, ' ', middlename) as 'Name'"
+ "FROM user "
+ "WHERE CAST(student_id as CHAR) LIKE '%" + tempSearchValue + "%'";
try {
pst = conn.prepareStatement(sql);
rs = pst.executeQuery();
while(rs.next()) {
table.setModel(DbUtils.resultSetToTableModel(rs));
empty = false;
}
if(empty) {
String error = "";
table.setModel(new javax.swing.table.DefaultTableModel(
new Object [][] {
{"No matching records found",null}
},
new String [] {
/** I WANT TO PUT THE SAME COLUMN NAME ON MY DATABASE SELECTED BUT DON't Know
WHAT FUNCTION TO DO*/
}
));
}
} catch (Exception e) {
JOptionPane.showMessageDialog(null, e.getMessage());
}
}
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我尝试这样但仍然给我NULL!这段代码在下面empty = false;
for(int i=0; i<table.getColumnCount(); i++) {
test[i] = table.getColumnName(i);
}
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Pra*_*ran 29
ResultSetMetaData metaData = resultSet.getMetaData();
int count = metaData.getColumnCount(); //number of column
String columnName[] = new String[count];
for (int i = 1; i <= count; i++)
{
columnName[i-1] = metaData.getColumnLabel(i);
System.out.println(columnName[i-1]);
}
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