作者st84514 ()
看板PHP
標題[請益] 用isset判斷是否為空的問題
時間Tue Apr 2 13:38:20 2013
錯誤訊息如下:
nWarning: mysql_fetch_array() expects parameter 1 to be resource, boolean
given in /var/www/html/android_login_api/include/reward.php on line
24nnWarning: mysql_num_rows() expects parameter 1 to be resource,.....
程式碼如下:
// check for post data
if (isset($_GET["M_ID"]) && isset($_GET["code"])) {
$M_ID = $_GET['M_ID'];
$code = $_GET['code'];
$result = mysql_query("SELECT sid,credit,img_url,name FROM survey_list
WHERE code = $code");
$row1 = mysql_fetch_array($result);
$credit = $row1["credit"];
$sid = $row1["sid"];
$img_url = $row1["img_url"];
$name = $row1["name"];
$check_record = mysql_query("SELECT * FROM record WHERE sid = $sid AND M_ID
= $M_ID");
$response["sid"] = $sid;
$response["M_ID"] = $M_ID;
//$response["check_record"] = mysql_num_rows($check_record);
//$response["result"] = mysql_num_rows($result);
if(mysql_num_rows($result) > 0 && mysql_num_rows($check_record) < 1) {
$user_result = mysql_query("SELECT credit FROM users WHERE M_ID = $M_ID");
$row2 = mysql_fetch_array($user_result);
$user_credit = $row2["credit"];
$Total_credit = $credit + $user_credit;
$update = mysql_query("UPDATE users SET credit = '$Total_credit' WHERE
M_ID = $M_ID");
$insert_record = mysql_query("INSERT INTO record(M_ID, sid, name, credit,
img_url) VALUES('$M_ID', '$sid', '$name', '$credit', '$img_url')");
if($update && $insert_record){
$response["success"] = 1;
$response["credit"] = $Total_credit;
}
}
else if(mysql_num_rows($result) > 0 && mysql_num_rows($check_record) > 0)//
已兌獎
{
$response["success"] = -1;
}
else{
$response["success"] = 0;
$response["message"] = "No products found";
}
echo json_encode($response);
}
我這邊用isset去判斷我手機使用者接收到的code值並判斷是否為空，
但現在bug是就算使用者不輸入任何值送出，也會出現成功訊息...
解很久解不出來...想麻煩版上前輩幫我看一下，當code值為空時，必須顯示
錯誤訊息，感激不盡，謝謝!
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1F：→ arrack:改成!=''如何？ 04/02 13:45
3F：→ st84514:您是說哪邊改成!='' 04/02 14:36
4F：→ st84514:map大，我將isset($_GET["code"])改成is null()還是錯誤.. 04/02 14:39
5F：→ st84514:解決了!感謝! 04/02 14:58
6F：→ MOONRAKER:有!empty()好用為何還要用 != '' 04/02 15:44