2014-11-01 57 views
-2

我有一個名稱產品的表之一它的列名稱是design_id其中設計ID是以逗號分隔格式保存像 - 5,4,6,51,85,65, 98,32,324,122,14,755,MySQL條款在哪裏條件不起作用

我一定要找到具有設計標識98而對於這一點,我已經寫了查詢

結果
SELECT * FROM `products` where `design_id` IN (98) 

查詢運行成功,但沒有返回任何結果?

列類型是varchar。

+2

了'IN'操作符周圍的其他方法 – 2014-11-01 11:21:46

+0

什麼樣的領域是什麼?整數或字符串與逗號分離? – SMA 2014-11-01 11:21:49

+0

請填寫完整代碼.. – Ali 2014-11-01 11:22:46

回答

1

例子:

<?php 
$con=mysqli_connect("localhost","my_user","my_password","my_db"); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

$sql="SELECT Lastname,Age FROM Persons ORDER BY Lastname"; 

if ($result=mysqli_query($con,$sql)) 
    { 
    // Fetch one and one row 
    while ($row=mysqli_fetch_row($result)) 
    { 
    printf ("%s (%s)\n",$row[0],$row[1]); 
    } 
    // Free result set 
    mysqli_free_result($result); 
} 

mysqli_close($con); 
?> 
1

IN運算符遍歷數據集,因此您必須反轉查詢。

SELECT * FROM table_name WHERE 98 IN column_name 

然而,在大多數數據庫中,平等的opterator要快得多。

SELECT * FROM table_name WHERE column_name=98 
+0

我不能使用這種類型的動態部分,我們不假定我們必須創建多少列。我的列結構是5,45,455,38,26,75,45, – 2014-11-01 11:33:53

+0

SELECT * FROM table_name WHERE 98 IN(5 ,45,455,38,26,75,45) – 2014-11-01 11:35:16

1

是對SELECT * FROM table_name where column_name='98'

更好地使用,但是要存儲:

嘗試使用INSERT

INSERT INTO table_name (column1,column2,column3,...) VALUES (value1,value2,value3,...);

不過是你想獲取:

嘗試使用mysql_fetch_row

例子:

<?php 
$con=mysqli_connect("localhost","my_user","my_password","my_db"); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

$sql="SELECT Lastname,Age FROM Persons ORDER BY Lastname"; 

if ($result=mysqli_query($con,$sql)) 
    { 
    // Fetch one and one row 
    while ($row=mysqli_fetch_row($result)) 
    { 
    printf ("%s (%s)\n",$row[0],$row[1]); 
    } 
    // Free result set 
    mysqli_free_result($result); 
} 

mysqli_close($con); 
?>