sql/php查询在phpmyadmin中工作，但在站点中不工作。
我注意到很多人都遇到过这个问题，但不可否认的是，我还没有这个网站上的一些程序员先进。=）我谦恭地请求您提供任何可能的经验：p谢谢。
<?php
// session_start();
// ob_start();
ini_set("display_errors", true);
error_reporting(-1);
// Connection to database.
mysql_connect('localhost', 'root', '') or die(mysql_error());
mysql_select_db('') or die(mysql_error());
?>
<?php
// Maintenance page.
$maintenance = 1; // 1 = Site Online || 0 = Site Offline
if ($maintenance == 0) {
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<head>
<style type="text/css">
body {
color:#ffffff;
background-color: #000000;
font-family: Arial, Helvetica, sans-serif;
}
</style>
</head>
<title></title>
</head>
<body>
<center><img src="images/p4flogo.png" /></center><br />
<?php
echo "<br/><br /><center>This site is currently under maintenance.</center>";
} else {
// Start of main website.
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<head>
<style type="text/css">
body {
color:#ffffff;
background-color: #000000;
font-family: Arial, Helvetica, sans-serif;
}
a:link {color: orange; text-decoration: underline; }
a:active {color: red; text-decoration: underline; }
a:visited {color: orange; text-decoration: underline; }
a:hover {color: red; text-decoration: none; }
table {
border-color: #333333;
}
th {
background-color:#ffffff;
color:#000000;
font-family: Arial, Helvetica, sans-serif;
height:30px;
}
td { font-family: Arial, Helvetica, sans-serif;
color:#ffffff;
height:35px;
}
</style>
</head>
<title></title>
</head>
<body>
<table border="0" cellspacing="1" align="center">
<tr><td>
<center><img src="images/p4flogo.png" /></center><br /><br />
<form action="" method="post">
Search for a soldier: <input type="text" name="value" size="35" /><input type="submit" value="search" /><br />
<?php
if (isset($_POST['value']) && !empty($_POST['value'])) {
$value = mysql_real_escape_string($_POST['value']);
// query to database for soldier stats
// query works in phpmyadmin but not on site.
$sql = "
SELECT
`Name`,
MAX(`Level`),
`Class`,
SUM(`Kills`),
SUM(`Deaths`),
SUM(`Points`),
SUM(`TotalTime`),
SUM(`TotalVisits`),
`CharacterID`
FROM
`Characters`
WHERE
`Name` LIKE '$value%' OR `CharacterID` LIKE '$value'
GROUP BY
`Name`,
`Class`,
`CharacterID`
ORDER BY
`Name` ASC;";
$query = mysql_query($sql);
$numrow = mysql_num_rows($query);
if ($numrow >= 1) {
echo "<br /><a href=\"index.php\"><b>View TOP 100 Players!</b></a><br />";
echo "<table border=\"1\" cellspacing=\"0\" cellpadding=\"5\"> ";
echo "<th>Soldier Name</th><th>Level</th><th>Class</th><th>KDR</th><th>Kills</th><th>Deaths</th><th>Points</th><th>Hours Played</th><th>Total Visits</th><th>CharacterID</th>";
echo "<br />";
WHILE ($row = mysql_fetch_assoc($query)) {
$SoldierName = $row['Name'];
$Level = $row['Level'];
$Class = $row['Class'];
if ($Class == NULL | empty($Class)) {
$Class = '<center>n / a</center>';
}
if ($Class == 1) {
$Class = 'Assault';
}
if ($Class == 2) {
$Class = 'Recon';
}
if ($Class == 3) {
$Class = 'Medic';
}
if ($Class == 4) {
$Class = 'Engineer';
}
echo $Kills = $row['Kills'];
if ($Kills == 0) {
$Kills = 1;
}
$Deaths = $row['Deaths'];
if ($Deaths == 0) {
$Deaths = 1;
}
$Kdr = round($Kills / $Deaths, 2);
$Points = $row['Points'];
$TimePlayed = $row['TotalTime'];
if ($TimePlayed == 0) {
$TimePlayed = 1;
} else {
$TimePlayed = round(($TimePlayed / 3600), 0);
}
$TotalVisits = $row['TotalVisits'];
$CharacterID = $row['CharacterID'];
echo "<tr>";
echo "<td><b>$SoldierName</b></td>";
echo "<td>$Level</td>";
echo "<td>$Class</td>";
if ($Kdr > 3.9) {
echo "<td><font color=\"red\"><b>$Kdr</b></font></td>";
} else if ($Kdr > 2.5 && $Kdr < 4) {
echo "<td><font color=\"orange\"><b>$Kdr</b></font></td>";
} else {
echo "<td><font color=\"limegreen\">$Kdr</font></td>";
}
echo "<td>$Kills</td>";
echo "<td>$Deaths</td>";
echo "<td>$Points</td>";
echo "<td>$TimePlayed</td>";
echo "<td>$TotalVisits</td>";
echo "<td>$CharacterID</td>";
echo "</tr>";
}
echo "</table>";
} else {
echo "No player found with that name. Please try again.";
}
} else {
if (empty($_POST['value'])) {
echo "<font color=\"red\">You must enter a search value.</font>";
}
// query to p4f database for top 100 players.
$sql = "SELECT * FROM `Characters` WHERE `Points` > 1 GROUP BY `Name` ORDER BY `Points` DESC LIMIT 100;";
$query = mysql_query($sql);
echo "<h3>TOP 100 PLAYERS</h3>";
echo "<table border=\"1\" cellspacing=\"0\" cellpadding=\"5\"> ";
echo "<th></th><th>Soldier Name</th><th>Level</th><th>Class</th><th>KDR</th><th>Kills</th><th>Deaths</th><th>Points</th><th>Hours Played</th><th>Total Visits</th><th>CharacterID</th>";
// echo "Made it to loop!";
$Rank = 1;
WHILE ($row = mysql_fetch_assoc($query)) {
$SoldierName = $row['Name'];
$Level = $row['Level'];
$Class = $row['Class'];
if ($Class == NULL | empty($Class)) {
$Class = '<center>n / a</center>';
}
if ($Class == 1) {
$Class = 'Assault';
}
if ($Class == 2) {
$Class = 'Recon';
}
if ($Class == 3) {
$Class = 'Medic';
}
if ($Class == 4) {
$Class = 'Engineer';
}
$Kills = $row['Kills'];
if ($Kills == 0) {
$Kills = 1;
}
$Deaths = $row['Deaths'];
if ($Deaths == 0) {
$Deaths = 1;
}
$Kdr = round($Kills / $Deaths, 2);
$Points = $row['Points'];
$TimePlayed = $row['TotalTime'];
if ($TimePlayed == 0) {
$TimePlayed = 1;
} else {
$TimePlayed = round(($TimePlayed / 3600), 0);
}
$TotalVisits = $row['TotalVisits'];
$CharacterID = $row['CharacterID'];
echo "<tr>";
echo "<td>$Rank</td>";
echo "<td><b>$SoldierName</b></td>";
echo "<td>$Level</td>";
echo "<td>$Class</td>";
if ($Kdr > 3.9) {
echo "<td><font color=\"red\"><b>$Kdr</b></font></td>";
} else if ($Kdr > 2.5 && $Kdr < 4) {
echo "<td><font color=\"orange\"><b>$Kdr</b></font></td>";
} else {
echo "<td><font color=\"limegreen\">$Kdr</font></td>";
}
echo "<td>$Kills</td>";
echo "<td>$Deaths</td>";
echo "<td>$Points</td>";
echo "<td>$TimePlayed</td>";
echo "<td>$TotalVisits</td>";
echo "<td>$CharacterID</td>";
echo "</tr>";
$Rank++;
}
echo "</table>";
}
}
?>
</td></tr>
</table>
</body>
</html>
最佳答案
我对mysql不太确定，但我知道在sql中，当您执行sum（Kills）之类的聚合函数时，就不能通过$row['kills'引用行。我不知道这是否是您的问题，但您可以尝试在select语句中将sum（Kills）作为“kills”。对聚合选择执行此操作将允许您一直引用它们。