2013-08-28 34 views
0

我正在使用ms sql server與php,我需要知道我做錯了什麼,因爲我沒有從數據庫中獲取任何數據。我知道我已經連線,並且我知道列是正確的名稱以及表單名稱。當我轉儲結果時,它說 (mssql結果)類型的資源(6)。手動查詢Ms sql

這是我的查詢。

$query = "SELECT * 
FROM [ERCOT_Data].[dbo].[TDSP_ESIID] 
WHERE ADDRESS LIKE '%$house_num%' 
    AND ADDRESS LIKE '%$street%' 
    AND ZIPCODE LIKE '%$zip%' 
    AND CITY LIKE '%$city%'"; 

我也試着像這樣

$query = "SELECT * 
FROM [ERCOT_Data].[dbo].[TDSP_ESIID] 
WHERE ADDRESS LIKE '$house_num%' 
    AND ADDRESS LIKE '$street%' 
    AND ZIPCODE LIKE '$zip%' 
    AND CITY LIKE '$city%' "; 
+0

運行通過你使用什麼軟件來管理你的數據庫查詢。首先確認數據是否與您的查詢相符 – andrewb

+0

您是否分析了查詢並直接在SQL Server上執行?你有什麼結果嗎?你如何指定參數值?例如$ house_num? – Nilesh

+0

是的,如果我手動從Sql manager studio中輸入數據,我會得到我想要的結果。我的查詢字符串有問題嗎? $ house_num是來自表單輸入的變量。 –

回答

0

試試這個!

這是更好地把PHP變量進行字符串這樣

$query = "SELECT * 
FROM [ERCOT_Data].[dbo].[TDSP_ESIID] 
WHERE ADDRESS LIKE '".$house_num."%' 
    AND ADDRESS LIKE '".$street."%' 
    AND ZIPCODE LIKE '$".zip."%' 
    AND CITY LIKE '$".city."%' "; 
0

試試這個

$query = "SELECT * 
FROM [php].[dbo].[TDSP_ESIID] 
WHERE ADDRESS LIKE '".$house_num%'." 
AND ADDRESS LIKE '".$street%'." 
AND ZIPCODE LIKE '".$zip%'." 
AND CITY LIKE '$city%' ";