PHP無法寫入mySQL
小妹是php新手
最近想要試著寫一個搜尋程式
主要是藉由php作為媒介連結android和mySQL
不過卻卡在我的php一直無法成功判讀SQL語法
以下是我主要問題程式片段:
(in java)
try {
String result = DBConnector.executeQuery("SELECT*FROM user where zip = 'newzip' ");
System.out.println(result);
(php語法)
$db = mysql_pconnect("localhost","root","1234");
mysql_query("SET CHARACTER SET 'UTF8';");
mysql_query("SET NAMES UTF8;");
mysql_query("SET CHARACTER_SET_CLIENT=UTF8;");
mysql_query("SET CHARACTER_SET_RESULTS=UTF8;");
mysql_select_db("food");
$sql = $_POST["query_string"];
$res = mysql_query($sql);
while($r = mysql_fetch_assoc($res))
$output[] = $r;
print(json_encode($output));
mysql_close();
?>
這是我的錯誤訊息:
mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\AndroidConnectDB\android_connect_db.php on line 13
我有將java中的語法直接在phpmyadmin中執行
(即SELECT*FROM user where zip = 'newzip')
是可以正確找到該筆符合條件的資料
但是不知道為什麼 透過java寫入至php中去呼叫mySQL執行時卻總是傳回錯誤
上網爬過很多文
有試著去修正過編碼的問題(SET NAMES UTF8;)
也有去php.ini修改過 magic_quotes_gpc = On
但仍舊回傳null以及錯誤訊息
但是我將它改成
SELECT*FROM user where id=1 時
是可以正確執行的00"
所以再猜想會不會是單引號引起的問題
也有將它改成
SELECT*FROM user where zip = /'newzip/'
卻仍舊不見好轉ˊˋ
想要請問各位大大是否有解決的方法?ˊˋ
------解决方案--------------------
猜测是没有用的,需要看到出错时的错误信息
$res = mysql_query($sql) or die(mysql_error());
因为你给出的错误信息是
mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource
即传递给 mysql_fetch_assoc 的不是有效的 MySQL 资源,所以要逐步向上查找
------解决方案--------------------
你不觉得应该是这样吗?
SELECT * FROM user where zip = 'newzip'
------解决方案--------------------
关闭魔术引号,修改php.ini magic_quotes_gpc = Off
或者临时对传入字符串反转义
$sql = stripslashes($_POST["query_string"]);也可以应付过来

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