2017-04-08 116 views
0

我有employeeinfo表和employeeloan表.. 和我使用此查詢:PHP的MySQL:連接兩個表

$result=mysqli_query($con,"select tblemployeeinfo.employeeid, tblemployeeloan.amount from tblemployeeinfo left join tblemployeeloan on tblemployeeinfo.employeeid = tblemployeeloan.employeeid where tblemployeeinfo.employeeid = <employee id for example employee id 1>"); 

while($row = mysqli_fetch_array($result)){ 
echo "<br>".$row['employeeid']; 
echo "<br>".$row['amount']; 
} 

結果是

employee id: 1 
amount: 1000 
employee id: 1 
amount: 500 
employee id: 1 
amount: 100 

,但我想要的是像顯示這個:

employee id: 1 
amount: 1000 
amount: 500 
amount: 100 

抱歉我的英文不好。 TNX提前

+1

你確定你的結果有員工ID?我沒有看到你在查詢中選擇了 – hassan

+0

對不起,先生...我的壞我會編輯我的帖子..據說這是tblemployeeid.employeeid –

+0

[左連接沒有重複行左表]的可能重複( http://stackoverflow.com/questions/22769641/left-join-without-duplicate-rows-from-left-table) –

回答

0

普通黑客可以通過下面的代碼來完成

$i = 0 ; 
while($row = mysqli_fetch_array($result)){ 
    if($i == 0) 
     echo "<br>".$row['employeeid']; 
    echo "<br>".$row['amount']; 
    $i++; 
} 
+0

謝謝先生..實際上它的工作..我現在將在我的代碼中使用它..但如果有其他方式,我很高興知道它..反正tnx再次先生.. –

+0

也可以檢查如果mysql可以做工作和解決這個問題 - > SELECT whatever_is_needed FROM table1 INNER JOIN table2 ON table1.otherid2 = table2.otherid2 GROUP BY whatever_id – OldPadawan