mysql和php在本地工作,而不是在godaddy上

我使用的是共享的Linux主机,我写的一些PHP在服务器上无法与我的数据库配合使用。

编辑:

我解决了这个问题。我不小心从配置文件中删除了“数据库”连接变量。真的。我真傻。

编辑:

不工作的意思是我看不懂数据库。我已经搞砸了,知道数据库正在运行,当我修改密码变量时,我得到了“拒绝访问”。此外,在本地主机上,密码是不同的,并且可以工作。

编辑2: 我现在正在使用一个示例脚本。现在它显示一个空白页。至少没有错误…我把密码改成字母数字。我没有使用“localhost”。

编辑3:使用正确的密码和服务器信息,我使用mysql_error()得到以下错误。我没有在数据库URL中使用前导http://。

无法通过socket'/var/lib/mysql/mysql.sock'连接到本地mysql服务器(2)

编辑4:

这是我的连接码,请注意 nl() 和 report() 是分别回送换行和诊断消息的预定义函数。连接变量位于上面包含的外部文件中。我查过了-包括服务都很好。

//////////////////////////////
// Connect to the database  /////////////////////////////////////////
//////////////////////////////

//report -  Attemting to establish connection ...
report(4);

////
//Step 1) Set connection in variable
////

$conn = mysql_connect($server, $user, $pass);

////
// Step 2) Verify the connection
////

if(!$conn){
    //report - failed, see next line...
    report(5);
    //report - FATAL ERROR: Check database name, username and password.
    report(9);      
}else{
  //report - done!
  report(7);
}

//report - Selecting database ...
report(6);

////
// Step 3) Select the database
////

$db_select = mysql_select_db($database);

////
// Step 4) Verify successful selection
////

if(!$db_select){
    //report - failed, see next line...
    report(5);
    //report - FATAL ERROR: Could not select database.
    report(8);  
}else{
  //report - done!
  report(7);
}

//report -  Running query ...
report(10);

////
// Step 5) Create the original the query
////

$selector = " * ";
$target = "`properties`";
$condition = "1";

$sql = "SELECT " . $selector . " FROM " . $target . " WHERE " . $condition;

////
//  Step 6) Check for a $_GET[] parameter
////

if($_GET['listing'] && !is_null($_GET['listing'])){                 //if a listing number was supplied
    $listingNum = htmlentities($_GET['listing']);                   //safety first - clean it from code injections                      
    $pattern = '/\b[0-9]{1,6}\b/';                                  //define a range of valid, sercheable listings
    if(preg_match($pattern,$listingNum) == 1){                      //if the listing id is a valid one
        $sql .= " AND `listing_id` =" .$listingNum . "";            //search for it in the database
    }elseif($exp == 0){                                             //if the listing number is out of range         
        if($listingNum != "all"){                                   //check if the "all" keyword was ommitted - if it was not supplied.
            //report - failure ... see below
            report(5);
            //report - Invalid listing ID
            report(12);
            //
            // Invalid Listing ID...
            //
        }       
    }
}else if(!$_GET['listing']){
    //report - failure ... see below
    report(5);
    //report - Invalid listing ID
    report(12);
    //
    //No listing ID
    //
}


if(!$sql){
    //report - failed, see next line...
    report(5);
    //report - FATAL ERROR: Could not run query.
    report(11);
    //
    // For some reason the query doesn't exist here. I really do wonder why.
    //
}else{
  //report - done!
  report(7);
  nl();
}

$result = mysql_query($sql);    //perform the actual query

if(!$result){
    echo("There's been some sort of error with the search lookup. Here are the details: \n" . mysql_error());
}

mysql_close($conn);         //close the connection to the SQL server