PHP while循环不在服务器上

Question

if ($area == Null || $area == "Area of City") { 
    $ment = "CREATE OR REPLACE VIEW water 
      AS SELECT postable.postid, description, dated, image, posterid, country, state, city, area, cat 
      FROM postable 
      INNER JOIN ".$_GET['Cat']." 
      ON postable.postid = ".$_GET['Cat'].".postid 
      WHERE ".$_GET["Cat"].".cat = '".$_GET["what"]."' 
      AND postable.country = '".$_GET["Country"]."' 
      AND postable.state = '".$_GET["State"]."' 
      AND postable.city = '".$_GET["City"]."'"; 
} 

$ment = "CREATE OR REPLACE VIEW water 
     AS SELECT postable.postid, description, dated, image, posterid, country, state, city, area, cat 
      FROM postable, ".$_GET["Cat"]." 
     WHERE (
      (postable.postid = ".$_GET["Cat"].".postid 
      AND ".$_GET["Cat"].".cat = '".$_GET["what"]."' 
      AND postable.country = '".$_GET["Country"]."' 
      AND postable.state = '".$_GET["State"]."' 
      AND postable.city = '".$_GET["City"]."' 
      AND postable.area = '".$area."') 
      OR 
      (postable.postid = ".$_GET["Cat"].".postid 
      AND ".$_GET["Cat"].".cat = '".$_GET["what"]."' 
      AND postable.country = '".$_GET["Country"]."' 
      AND postable.state = '".$_GET["State"]."' 
      AND postable.city = '".$_GET["City"]."') 
     )"; 
$tester = "SELECT * FROM water, userguy 
      WHERE userguy.posterid = water.posterid"; 

if (!mysql_query($ment, $con)) { 
    die('There are no posts matching your search, please enter another search in either another location or category, or both. '); 
} 

if (!mysql_query($tester,$con)) { 
    die('There are no posts matching your search, please enter another search in either another location or category, or both.'); 
} 

$result = mysql_query($tester,$con); 


while ($row = mysql_fetch_array($result)) { 
if ($row == Null) { 
    echo "<tr><td colspan'4'>There are no posts matching your search, please enter another search in either another location or category, or both.</td></tr>";} 
    echo "<tr><td colspan='4' class='sult'>"; 
    echo "<div id='main'>".$row['description']."</div> ".$row['dated']." <a  href='mai.php?tofield=".$row['email']."'><b>".$row['email']."</b></a><b>&nbsp".$row['mobile']."</b>"; 
    if ($row['image'] != Null) { 
    echo "<img src='upload/".$row['image']."' class='relt'/>"; 
    } 
    echo "</td></tr>"; 
} 

?> 

<tr> 
    <td colspan='4'> 
    <br/><br/><br/><br/> 
    </td> 
</tr> 
<tr> 
    <td colspan='2' align='center'> 
    <img src='first.jpg'/> 
    </td> 
    <td colspan='2' align='center'> 
    <img src='second.jpg'/> 
    </td> 
</tr> 
<tr> 
    <td colspan='2' align='center'> 
    <img src='high.jpg'/> 
    </td> 
    <td colspan='2'></td> 
</tr> 
</table> 

</body> 

好返回任何东西，现在，这个代码工作在我的电脑，我用它来开发这个我的网站上，这是它从数据库返回的行。但是，当我将这段代码放到我的网站上时，它不会返回任何内容，实际上，脚本甚至不显示任何内容。我会很感激，如果我可以得到任何帮助，它现在已经让我不眠之夜太久了...

Answer 1

我推测从$_GET用法，你在浏览器中运行这个，而不是从一个控制台。这可能是因为你的mysql扩展没有加载，或者你没有与数据库服务器的有效连接。

• 首先，检查你的错误日志。这些通常与您的Apache访问日志*保持在同一个地方。这会给你一个错误信息来处理。
• 接下来，在您服务器上的空白页面中运行<?php phpinfo(); ?>，并查看您的基于Web的PHP是否支持MySQL支持。
• 尝试运行mysql_error()后可能会失败的数据库操作。这也是挖掘错误的好方法。

最后要记住，你的脚本目前有SQL注入安全漏洞。修复这些probs之前，你去住这个:-)。

编辑：*您的服务器日志的位置取决于您正在使用的操作系统，并且您没有指定此位置。这很容易从搜索引擎中获得，但是 - 如果您在Windows上，那么请执行“Windows apache日志位置”的网页搜索。简单;-)

编辑2：如果SSH已被禁用，并且您觉得您需要它，请与您的主机通话。然而，这是一个单独的问题，你不需要它来解决手头的问题。如果你的phpMyAdmin有问题，请重新安装它，或使用你的主机提供的那个。 （如果您使用的是cPanel，则包含在您的控制面板中。）

Answer 2

我看不到任何直接的原因，为什么您的脚本不应该运行，因为halfer指出您的消息和错误日志是最好的地方看。

顺便说一句，虽然：在while循环的第一行if($row = Null)永远不会为真（如果它是，while循环将不进入） 考虑

$n_rows = 0; 
while($row = mysql_fetch_array($result)) { 
    $n_rows++; 
    // Process row 
} 
if(!$n_rows) { 
    // No rows message 
} 

（我知道这应该是评论不是一个答案，但我没有点）

Answer 3

试试这个（见注释代码更改）：

<?php 

    if ($area == Null || $area == "Area of City") { 
    $ment = "CREATE OR REPLACE VIEW water 
      AS SELECT postable.postid, description, dated, image, posterid, country, state, city, area, cat 
       FROM postable 
      INNER JOIN ".$_GET['Cat']." 
       ON postable.postid = ".$_GET['Cat'].".postid 
      WHERE ".$_GET["Cat"].".cat = '".$_GET["what"]."' 
       AND postable.country = '".$_GET["Country"]."' 
       AND postable.state = '".$_GET["State"]."' 
       AND postable.city = '".$_GET["City"]."'"; 
    } else { 
    // Pretty sure this should be in an else block 
    // As it was, the query above would never be executed, because it would 
    // always be overwritten by this one 
    $ment = "CREATE OR REPLACE VIEW water 
      AS SELECT postable.postid, description, dated, image, posterid, country, state, city, area, cat 
       FROM postable, ".$_GET["Cat"]." 
      WHERE (
       (postable.postid = ".$_GET["Cat"].".postid 
       AND ".$_GET["Cat"].".cat = '".$_GET["what"]."' 
       AND postable.country = '".$_GET["Country"]."' 
       AND postable.state = '".$_GET["State"]."' 
       AND postable.city = '".$_GET["City"]."' 
       AND postable.area = '".$area."') 
       OR 
       (postable.postid = ".$_GET["Cat"].".postid 
       AND ".$_GET["Cat"].".cat = '".$_GET["what"]."' 
       AND postable.country = '".$_GET["Country"]."' 
       AND postable.state = '".$_GET["State"]."' 
       AND postable.city = '".$_GET["City"]."') 
      )"; 
    } 

    $tester = "SELECT * FROM water, userguy 
      WHERE userguy.posterid = water.posterid"; 

    /* 
    Do you really not want the result of this query? 
    Shouldn't you be catching the results in a variable? 
    If not, consider adding a LIMIT 1 clause to the query, as the 
    database will have to do more work to return a full result set 
    you never use... 
    Regardless of that, testing for !mysql_query doesn't tell you there 
    were no results, it tells you whether the query itself failed. You 
    test the number of results using mysql_num_rows() or a SELECT count() query 
    */ 
    if (mysql_num_rows(mysql_query($ment, $con)) < 1) { 
    die('There are no posts matching your search, please enter another search in either another location or category, or both. '); 
    } 

    // Same goes for this. 
    // However, since you definitely do want the results of this, why run it twice? 
    $result = mysql_query($tester,$con); 
    if (mysql_num_rows($result) < 1) { 
    die('There are no posts matching your search, please enter another search in either another location or category, or both.'); 
    } 

    // Use mysql_fetch_assoc() instead of mysql_fetch_array(), it's more efficient 
    while ($row = mysql_fetch_assoc($result)) { 
    // if ($row == Null) { 
    // Sorry, what? If the row is null?? 
    // It will never be null... if it were, the loop would break immediately 
    // and never reach this point, plus mysql_fetch_* never returns null! 
    // echo "<tr><td colspan'4'>There are no posts matching your search, please enter another search in either another location or category, or both.</td></tr>"; 
    // } 
    echo "<tr><td colspan='4' class='sult'>"; 
    echo "<div id='main'>".$row['description']."</div> ".$row['dated']." <a  href='mai.php?tofield=".$row['email']."'><b>".$row['email']."</b></a><b>&nbsp".$row['mobile']."</b>"; 
    if ($row['image'] != Null) { 
     echo "<img src='upload/".$row['image']."' class='relt'/>"; 
    } 
    echo "</td></tr>"; 
    } 

?> 

<tr> 
    <td colspan='4'> 
    <!-- 
     Really? You can't just use a CSS height? 
    --> 
    <br/><br/><br/><br/> 
    </td> 
</tr> 
<tr> 
    <td colspan='2' align='center'> 
    <img src='first.jpg'/> 
    </td> 
    <td colspan='2' align='center'> 
    <img src='second.jpg'/> 
    </td> 
</tr> 
<tr> 
    <td colspan='2' align='center'> 
    <img src='high.jpg'/> 
    </td> 
    <td colspan='2'></td> 
</tr> 
</table> 

</body>