我发挥了职能,让我了解用户名称。 不同的用户先令将同时使用,这样功能将多次使用。 它对一名用户进行罚款,但如果有一个以上用户,则给我一个错误。 错误是“战争:我sql_select_db():提供的论点不是有效的MySQL-Link资源”。
法典
<?php
function user_details($user_id) {
require_once( ../Connections/runner.php );
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "")
{
$theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
$theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string ($theValue) : mysql_escape_string($theValue);
switch ($theType) {
case "text":
$theValue = ($theValue != "") ? " " . $theValue . " " : "NULL";
break;
case "long":
case "int":
$theValue = ($theValue != "") ? intval($theValue) : "NULL";
break;
case "double":
$theValue = ($theValue != "") ? " " . doubleval($theValue) . " " : "NULL";
break;
case "date":
$theValue = ($theValue != "") ? " " . $theValue . " " : "NULL";
break;
case "defined":
$theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
break;
}
return $theValue;
}
}
$colname_users = $user_id;
$first_name = "";
$last_name = "";
mysql_select_db($database_runner, $runner);
$query_users = sprintf("SELECT first_name, last_name, profile_img_small FROM sign_up WHERE user_id = %s", GetSQLValueString($colname_users, "int"));
$users = mysql_query($query_users, $runner) or die(mysql_error());
$row_users = mysql_fetch_assoc($users);
$totalRows_users = mysql_num_rows($users);
$first_name = $row_users[ first_name ];
$last_name = $row_users[ last_name ];
$profile_sml = $row_users[ profile_img_small ];
echo "$first_name $last_name";
}
?>
<?php
$user_id = 10;
?>
<a href="*"><?php user_details("$user_id"); ?></a>
<?php
$user_id = 9;
?>
<a href="*"><?php user_details("$user_id"); ?></a>