Question

我正在尝试在PHP页面上显示MySQL表。表头正确显示，但没有出现MySQL表行，并且出现MySQL表行数据时也会出现错误。我在运行PHP页面时收到此错误：

警告：mysqli_fetch_row（）要求参数1为mysqli_result，第61行/home/www/s.com/yogaclub/index.php中给出布尔值

这是我页面的代码：

   <?php
$username = "d";
$password = "dar";
$hostname = "localhost"; 

//connection to the database
$con = mysqli_connect($hostname, $username, $password) 
  or die("Unable to connect to MySQL");

if ($con) { } else { echo "didn't work";} 
?>

<!doctype html>
<head>
<style>
body {
    font-family:verdana;
}

#box {
    background-color:yellow;
    position:relative;
    color:red;
    padding-top:10px;
    margin:auto;
    width:600px;
    border-radius:15px
    padding-bottom:10px;    
}

    .container {
        width: 500px;
        clear: both;
    }
    .container input {
        width: 100%;
        clear: both;
    }


#pic {
    align:left
}
</style>    
<title>Yoga Club</title>
</head>
<body bgcolor="blue">
<div id="box">
<h3><center>Yoga Club!</center></h3>
<img src="thOX7T2TIR.jpg" id="pic"><span style="text-align:center; align:right"><span style="color:blue;"></span>
<?php 

$result = mysqli_query($con, "SELECT * FROM yogaclub");

echo "<table border='1'>
<tr>
<th>First Name</th>
<th>Last Name</th>
</tr>";

while($row = mysqli_fetch_row($result))
{
echo "<tr>";
echo "<td>" . $row['fname'] . "</td>";
echo "<td>" . $row['lname'] . "</td>";
echo "</tr>";
}
echo "</table>";

mysqli_close($con);
?><br /><br /><center>Please Sign up to the Adventures of Yoga.
We will be doing lots of different kinds of stretches.</center>
<?php /* if(isset($_POST['add']))
{ 
$sql = mysql_query("insert into yogaclub(fname, lname) values('$_GET['fname']', '$_GET['lname']')"); 
} else { echo "nonsend"; } */
?><br /><form method="post" action="<?php $_PHP_SELF ?>">First Name: <input name="fname" type="text"><Br />Last Name: <input type="text"><br />Address: <input type="text"><br />Email: <input type="text"><br />Phone Number: <input type="text"><input type="hidden" value="send" name="send"><br />Yoga Experience: <select>
   <option value="none">None</option>
   <option value="alittle">A little</option>
   <option value="alot">A lot</option>
</select> <br />Age: <input type="text" size="3"><br /><br />
<input type="submit" name="add" value="Sign me up!"></form></span>
</div>
</body>
</html>

非常感谢所有帮助。

Answer 1

测试$ result以验证查询至少返回一行可能是一个好主意。

Answer 2

而不是

while($row = mysqli_fetch_row($result))

请尝试使用

进行测试

while($row= mysqli_fetch_assoc($result))

尝试在HTML表中显示MySQL表数据时出现错误

2 个答案: