无法在简单的php-mysql查询中找到错误
[英]not able to find the error in a simple php-mysql query
问题描述
in this siomple code where i have to retrieve from table personnel which is having a column fname and some other columns as well , im not able to retrieve by matching fname. 
在这个简单的代码中，我必须从具有fname列和其他一些列的表人员中检索，我无法通过匹配fname进行检索。 code:-> 代码：->
<!DOCTYPE html>
<html>
<title>fname searched!!!</title>
<body>
<div id="container" style="width:1500px"> <! div for main header ... orange portion>
<div id="header" style="background-color:#FFA500;">
<h1 style="margin-bottom:0;">LIST OF SEARCHED ACTIVE SERVICE PERSONNEL</h1>
<figure>
<img src="personnel_head.jpg" alt="botg" width="1400" height="250">
<figcaption>.....tracking records......</figcaption>
</figure>
</div>
<div id="menu" style="background-color:#FFD700;height:1050px;width:200px;float:left;">
<! div for left side menus ... yellow portion>
candidate_reg_form<br>
view enrolled persons<br>
list of ammunitions<br>
character_certificate<br>
list_dependents<br>
regiments<br>
current serving general<br>
<b>SEARCH<b><br>
</div>
<! div for main container ... grey portion>
<div id="content" style="background-color:#EEEEEE;height:1050px;width:1300px;float:left;">
<BODY>
<fieldset>
<h1>active service personnel's details=>></h1>
<?php
$con=mysqli_connect("localhost","sumit","","ind_army");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$fname2=$_POST['fname2'];
echo"<br>";
$result = mysqli_query($con,"SELECT * FROM 'personnel' WHERE fname=$fname2");
echo "<table border='1'>
<tr>
<th>FNAME</th>
<th>MNAME</th>
<th>LNAME</th>
<th>SERV_NO</th>
<th>SEX</th>
<th>RANK</th>
<th>BDATE       </th>
<th>R_NAME</th>
<th>SALARY</th>
<th>ADDRESS</th>
<th>SUPER_SERV_NO</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['fname'] . "</td>";
echo "<td>" . $row['mname'] . "</td>";
echo "<td>" . $row['lname'] . "</td>";
echo "<td>" . $row['SERV_NO'] . "</td>";
echo "<td>" . $row['sex'] . "</td>";
echo "<td>" . $row['RANK'] . "</td>";
echo "<td>" . $row['bdate'] ."</td>";
echo "<td>" . $row['R_NAME'] . "</td>";
echo "<td>" . $row['salary'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['super_serv_no'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
<br><br>
<font color="orange">enter the service no for knowing his/her service ` `details</font>
<!-- form to get key detail of personnel record in database -->
<form name="form" method="POST" action="http://localhost/search_per2.php">
serv_no<input type="text" name="search"> <br><br>
<input type="submit" value="submit">
</form>
<font color="blue">enter the service no for knowing his/her depedents</font>
<!-- form to get key detail of dependent record in database -->
<form name="form" method="POST" action="http://localhost/dep_search.php">
serv_no<input type="text" name="search2"> <br><br>
<input type="submit" value="submit">
</form>
<LABEL>
<form action="http://localhost/indarmy.php" method="post" >
<form>
<input type="SUBMIT" name="HOME" VALUE="HOME"></label>
</FORM>
</FIELDSET>
<div id="footer" style="background-color:#FFA500;clear:both;text-align:center;">
Copyright © indarmy.com</div>
</div>
</body>
</html>
and the error it is giving is:--> Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\\xampp\\htdocs\\search_fname.php on line 66 它给出的错误是：->警告：mysqli_fetch_array（）期望参数1为mysqli_result，在66行的C：\\ xampp \\ htdocs \\ search_fname.php中给出布尔值
4 个解决方案
解决方案1
0 2014-04-23 14:07:29
Change 更改
while($row = mysqli_fetch_array($result))
to: 至：
while($row = mysqli_fetch_assoc($result))
If you loop fetch the result as an array you have loop through the array, whereas getting the row as associated array you will be able to reference the columns as you are doing now in your loop. 如果循环获取结果作为数组你通过数组有循环，而得到行作为关联数组，你将能够引用列，你在你的循环，现在做的事情。
解决方案2
0 2014-04-23 14:23:02
To see what's the error put: 要查看发生了什么错误：
if(!$result)
{
echo "error: ".mysqli_error($con);
}
right after the query execution. 在查询执行之后。
It's probably something in the SQL. 这可能是SQL中的东西。
Consider also mysqli_error_list 还考虑mysqli_error_list
解决方案3
0 2014-04-23 14:30:53
One you think you can definitely try is cleaning up your query just a tad bit by adding quotes around your search string $fname. 您认为可以肯定尝试的一种方法是，通过在搜索字符串$ fname周围添加引号来清理查询。 Im pretty sure cant use single quotes, so i replaced those with backticks. 我很确定不能使用单引号，所以我用反引号代替了它们。
$result = mysqli_query($con,"SELECT * FROM `personnel` WHERE fname = '$fname2' ");
解决方案4
-1 已采纳 2014-04-23 14:01:37
尝试如下查询：
$result = mysqli_query($con,"SELECT * FROM `personnel` WHERE fname='$fname2'");
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