我的脚本在我的本地工作，而不是托管

互联网上的工作我得到了托管在网络上这样的警告，页面上的报告（小时-male.php）如下我的脚本在我的本地工作，而不是托管

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/hatimmak/public_html/ssp12/report-hours-male.php on line 28

我的问题是：的脚本工作正常我的localhost >>在我的电脑上虽然不工作在互联网托管!!!!!

这是页面

<?php 


include 'config.php'; 

?><BR> 

Report for year <?php echo $_SESSION['year']; ?> semester <?php echo $_SESSION['sem']; ?> 

<?php 
// list of teachers Table rows and total hours 

echo "<table border='0' id='hor-minimalist-b' > 
<tr> 
<th scope='col'>Instructor name</th> 
<th scope='col'>Total hours for this semester</th> 
</tr>"; 


     //$gender = $_SESSION['gender']; 
     $year = $_SESSION['year']; 
     $sem = $_SESSION['sem']; 

     $tab = "Report$year$sem"; 
     $query = "SELECT teacher, SUM(hours) FROM $tab GROUP BY teacher"; 
     $result = mysql_query($query) ; 



while($row = mysql_fetch_array($result)) 
{ 


     $time = gmdate(DATE_RFC822); 

     //$gender = $_SESSION['gender']; 
     $year = $_SESSION['year']; 
     $sem = $_SESSION['sem']; 
     //$teacher = $_row['teacher']; 
     //$sumhours = $_row['SUM(hours)']; 



echo "<tr>"; 
echo "<td> ". $row['teacher']." </td>"; 
echo "<td> ". $row['SUM(hours)']." </td>"; 

echo "</tr>"; 
} 
echo "</table>"; 

?> 

<!--chart--> 
<html> 
    <head> 
    <!--Load the AJAX API--> 
    <script type="text/javascript" src="https://www.google.com/jsapi"></script> 
    <script type="text/javascript"> 

     // Load the Visualization API and the piechart package. 
     google.load('visualization', '1.0', {'packages':['corechart']}); 

     // Set a callback to run when the Google Visualization API is loaded. 
     google.setOnLoadCallback(drawChart); 

     // Callback that creates and populates a data table, 
     // instantiates the pie chart, passes in the data and 
     // draws it. 
     function drawChart() { 

     // Create the data table. 
     var data = new google.visualization.DataTable(); 
     data.addColumn('string', 'Topping'); 
     data.addColumn('number', 'hours'); 
     data.addRows([ 

<?php 


     $queryr = "SELECT teacher, SUM(hours) FROM $tab GROUP BY teacher"; 
     $resultr = mysql_query($queryr) ; 

while($rowr = mysql_fetch_array($resultr)) 
{ 
     $teacherr = $_rowr['teacher']; 
     $sumhoursr = $_rowr['SUM(hours)']; 


echo "['".$rowr['teacher']."' , ".$rowr['SUM(hours)']."]," ; 
} 

?> 
      ['', 0] 


     ]); 

     // Set chart options 

     var options = {'title':'Contact hours/instructor', 
       'legend':'left', 
       'is3D':true, 

       'width':800, 
       'height':300}; 

     // Instantiate and draw our chart, passing in some options. 
     var chart = new google.visualization.BarChart(document.getElementById('chart_div')); 
     chart.draw(data, options); 
     } 
    </script> 

    </head> 

    <body> 
    <!--Div that will hold the pie chart--> 
    <center><div id="chart_div"></div></center> 

    <form> 
<input type="button" value="Print This Report" onclick="window.print();"> </form> 

    </body> 
</html> 

<?php 

ob_end_flush(); 

?>

这是个config.php文件

<?php 

$con = mysql_connect("localhost","hatimmak_ssp","userpass"); 
if (!$con) 
    { 
    die('Could not connect: ' . mysql_error()); 
    } 

$sel = mysql_select_db("hatimmak_ssp", $con) or die(); 

?>

请帮助我:(

来源

2012-05-11 Hatim

数据库资源在哪里？ – k102

显示你的'config.php' – Norse

你的数据库连接可能没有成功。 – halfer

我怀疑是这些中的任何一个的价值：$ tab，teacher，hours。
他们需要完全匹配您的主机。

$tab = "Report$year$sem"; 
    $query = "SELECT teacher, SUM(hours) FROM $tab GROUP BY teacher";

来源

2012-05-11 08:11:39 adydy

谢谢，解决了，问题出在$ tab =“Report $ year $ sem”; 它应该是$ tab =“report $ year $ sem”; 只有R >> r 非常感谢您对Mr.adydy :) – Hatim

@欢迎您的光临，欢呼:) – adydy

-1

确定查询成功，您应该检查是否有错误？

if (mysql_query($sql)) 
{ 
    // your code 
} 
else 
{ 
    die(mysql_error()); 
}

来源

2012-05-11 07:48:01 Julien

我这样做了，我得到了这个错误 – Hatim

这在你的代码中并不明显。正如Carlos提到的那样，您应该只包含相关代码以及db连接所在的代码。如果这段代码在你的测试系统上正常工作，并且你的数据库在测试系统和产品系统上都是一样的话，那就意味着我能想到的唯一问题就是MySQL连接问题。 – Julien

我改变了$结果Gokhan Ozturk说，我得到了这个问题，报告年学期错误：表'hatimmak_ssp.Report'不存在 – Hatim

我的脚本在我的本地工作，而不是托管

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