Question

我尝试使用单选按钮来控制我的mysql查询，但是我遇到了一些初学者的障碍。我要么犯了小错误，要么设置错误。

（1）我通过使用数组将单选按钮的显示值设置为查询A的结果。这很好。

（2）我还将数组值设置为该单选按钮的值，而所有单选按钮都具有相同的名称。我假设这将允许单选按钮名称作为变量传递到查询B，但是我将单选按钮名称（'rbreed'）设置为等于变量然后使用变量（$ choice））在查询B中。

我注意到，当我第一次查看页面时，默认情况下不会检查任何按钮，并且可能是变量为null，因此查询为空，页面不起作用。此外，如果我单击其中一个按钮，它仍然不会产生任何结果，并且所选按钮将被取消选中。

我已经创建了一个if else语句来处理这个并略微更改查询，但我坚持使用“警告：mysqli_fetch_array（）期望参数1为mysqli_result，在C：\ wamp \ www \ NS \中给出布尔值第59行的view.php“

这是我的代码

<?php

$con = mysqli_connect("localhost","Nibbs","password");

if (!$con) {
    die ("You have a connect error: " . mysqli_connect_error());
}
mysqli_select_db($con,"Dogs");

$choice = 'rbreed';

if ($choice = '') {
    $sql = "SELECT * FROM register";
    $myData = mysqli_query($con,$sql);
    mysqli_query($con,$sql);
} else {
    $sql = "SELECT * FROM register WHERE hounds = $choice";
    $myData = mysqli_query($con,$sql);
    mysqli_query($con,$sql);
}

$radiosql = "SELECT DISTINCT Breed FROM register";
$myRData = mysqli_query($con,$radiosql);
$myRarray = array();

while ($row = mysqli_fetch_array($myRData,MYSQL_ASSOC)){
  $myRarray[] = $row;
}

echo "Select your type of hound:";
echo "<br />";
echo "<br />";
echo "<form action='' method='post'>";
echo "<input type='radio' name='rbreed' value=''>All";
echo "<input type='radio' name='rbreed' value=" . $myRarray[0]['Breed'] . ">" . $myRarray[0]['Breed'];
echo "<input type='radio' name='rbreed' value=" . $myRarray[1]['Breed'] . ">" . $myRarray[1]['Breed'];
echo "<input type='radio' name='rbreed' value=" . $myRarray[2]['Breed'] . ">" . $myRarray[2]['Breed'];
echo " "."<input type='submit' name='submit' value='Select' />";
echo "</form>";

echo "<br />";
echo "<table border=1>
<tr>
<th>Register ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Breed</th>
<th>Weight</th>
<th>Age</th>
<th>Sex</th>
</tr>";

while($record=mysqli_fetch_array($myData)){
    echo "<tr>";
    echo "<td><input type='text' name='reg_id' value='" . $record['reg_id'] . "'/> </td>";
    echo "<td><input type='text' name='first_name' value='" . $record['First_Name'] . "'/> </td>";
    echo "<td><input type='text' name='last_name' value='" . $record['Last_Name'] . "'/> </td>";
    echo "<td><input type='text' name='breed' value='" . $record['Breed'] . "'/> </td>";
    echo "<td><input type='int' name='weight' value='" . $record['Weight'] . "'/> </td>";
    echo "<td><input type='int' name='age' value='" . $record['Age'] . "'/> </td>";
    echo "<td><input type='text' name='sex' value='" . $record['Sex'] . "'/> </td>";
    echo "</tr>";
}

Answer 1

更改

 $sql = "SELECT * FROM register WHERE hounds = $choice";

到

 $sql = "SELECT * FROM register WHERE hounds = '". mysqli_real_escape_string($choice). "'";

否则，这是一个SQL语法错误。

Answer 2

首先改变

if ($choice = '') {
    $sql = "SELECT * FROM register";
    $myData = mysqli_query($con,$sql);
    mysqli_query($con,$sql);
}

进入

if ($choice == '') { 
    $sql = "SELECT * FROM register";
    $myData = mysqli_query($con,$sql);
    mysqli_query($con,$sql);
}

在while循环之前，你必须检查num_rows＆gt; 0

$rowcount=mysqli_num_rows($myData);
 if($rowcount > 0 )
  {
while ($row = mysqli_fetch_array($myRData,MYSQL_ASSOC)){
  $myRarray[] = $row;
}

}

将单选按钮值传递给mysql查询

2 个答案: