的mysql_query（）預計參數1爲字符串，資源給予

-3

$abc = mysql_fetch_array($result); 
    if (!$result) 
    die("Error: Data not found..");   

$name=$abc['name'] ; 
$email= $abc['email'] ;     


$sql = "UPDATE example SET name ='$name', email ='$email' WHERE id = '$id'"; 
    if(mysql_query($con, $sql)) //Error 
     Echo "Record Update successfully"; 
    else 
     Echo "ERROR: could not able to execute $sql".mysql_error($con); 

mysql_close($con);

來源

2016-01-09 Vrajesh Savaliya

mysql_query（）期望參數1是字符串，給定的資源//這是錯誤 –

'mysql_query（$ sql，$ con）' – smoqadam

1錯誤解決thnks，＆另一個錯誤=無法執行更新示例SET name =''，email =''WHERE id ='19'Table'test.example'does not exist –

更新你的代碼是這樣的：

$sql = "UPDATE example SET name = $name,email = $email WHERE id = $id"; 
mysql_query($sql,$con);

來源

2016-01-09 11:20:41 RK12

您需要交換$ CON和$闕paramters在你的mysql_query函數。

http://php.net/manual/en/function.mysql-query.php

來源

2016-01-09 11:22:51

你必須以$ CON

if(mysql_query($sql, $con)) 
     { 
     Echo "Record Update successfully"; 
     } 
     else 
     { 
     Echo "ERROR: could not able to execute $sql".mysql_error($con); 
     }

來源

2016-01-09 11:25:05

這個問題的答案已經給了很多次，請參考以下鏈接替換$ SQL。

Warning: mysql_query() expects parameter 1 to be string,

Error PHP> Warning: mysql_query() expects parameter 1 to be string, resource given in C:\xampp\htdocs\test1\Database.php on line 29

mysql_query() expects parameter 1 to be string, resource given

而且如果不能解析那麼問題走在這條鏈接。

https://stackoverflow.com/help/duplicates

https://stackoverflow.com/help/how-to-ask

來源

2016-01-09 11:28:48