Question

我收到未定义索引的错误，原因是什么？

while ($row = mysqli_fetch_array($result)) {
        echo "<tr>";

        echo "<td>" . $row['CandidateName'] . "</td>";
        echo "<td>" . $row['Position'] . "</td>";
         echo "<td><input type='radio' name='candidateid' value='".$row['candidateID']."' >";
      echo "<td>" . $row['NumberofVotes'] . "</td>";
        $candidateid=$row['CandidateID'];

    }

这是错误

Array ( [0] => 1 [CandidateID] => 1 [1] => Jejomar Binay [CandidateName] => Jejomar Binay [2] => President [Position] => President [3] => [NumberofVotes] => ) Array ( [0] => 2 [CandidateID] => 2 [1] => Mar Roxas [CandidateName] => Mar Roxas [2] => President [Position] => President [3] => 1 [NumberofVotes] => 1 )

我现在将向您展示整个代码及其工作现在，我的输出是在单选按钮工作时添加1个投票数。它没有错误，但是当我选择第一个单选按钮时，它只更新第二个数据。

<html>
        <center>
        <font size="2" face = "century gothic">
        <?php
        $con=mysqli_connect("localhost","root","","election2016");
        // Check connection
        if (mysqli_connect_errno())
        {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
        $result = mysqli_query($con,"SELECT * FROM candidate_info");
        echo "<table border='1'>
        <tr>
        <th>Candidate Name</th>
        <th>Position</th>
        <th>Vote</th>
        <th>Number of Votes</th>
        </tr>";
        while ($row = mysqli_fetch_array($result)) {
            echo "<tr>";

            echo "<td>" . $row['CandidateName'] . "</td>";
            echo "<td>" . $row['Position'] . "</td>";
             echo "<td><input type='radio' name='candidateid' value='".$row['CandidateID']."' >";
          echo "<td>" . $row['NumberofVotes'] . "</td>";
            $candidateID=$row['CandidateID'];

        }
        echo "</table>";
        mysqli_close($con);
        ?>

        <br>
        <br>
        <form method = "post" action = "<?php $_PHP_SELF ?>">
     <input type="text" name="candidateid" value="<?php echo $candidateID;?>">
         <input name = "update" type = "submit" id = "update" value = "update">
        </form>
        </center>
        </font>
        </html>
        <?php
                 if(isset($_POST['update'])) {
                    $dbhost = 'localhost';
                    $dbuser = 'root';
                    $dbpass = '';
 $candidateid = $_POST['candidateid'];
                    $conn = mysql_connect($dbhost, $dbuser, $dbpass);

                    if(! $conn ) {
                       die('Could not connect: ' . mysql_error());
                    }

                    $candidateid = $_POST['candidateid'];


                    $sql = "UPDATE candidate_info SET numberofvotes = 1 WHERE candidateid = '$candidateid'" ;
                    mysql_select_db('election2016');
                    $retval = mysql_query( $sql, $conn );

                    if(! $retval ) {
                       die('Could not update data: ' . mysql_error());
                    }
                    echo "Updated data successfully\n";

                    mysql_close($conn);
                 }
        ?>

Answer 1

您的数组$row中没有关键字'candidateid'请var_dump($row);查看关键名称，或者这些名称与数据库表格中的列名称相同，检查它的名称。

Answer 2

该字段＆＃34; candidateid＆＃34;应该是整数数据类型，但在更新查询中使用＆＃39;＆＃39;（单引号）包含此字段值？

$sql = "UPDATE candidate_info SET numberofvotes = 1 WHERE candidateid = '$candidateid'";

如果是整数数据类型，则应删除单引号

$sql = "UPDATE candidate_info SET numberofvotes = 1 WHERE candidateid = $candidateid";

并且在MySQL中，每个字段名称都区分大小写，因此您告诉字段名称是

候选人，候选人姓名，职位，号码投票

所以，你也应该在检索值时使用这些名称

    <?php
             if(isset($_POST['update'])) {
                $dbhost = 'localhost';
                $dbuser = 'root';
                $dbpass = '';
                $candidateid = $_POST['candidateid'];
                $conn = mysql_connect($dbhost, $dbuser, $dbpass);

                if(! $conn ) {
                   die('Could not connect: ' . mysql_error());
                }

                $candidateid = $_POST['candidateid'];


                $sql = "UPDATE candidate_info SET numberofvotes = numberofvotes + 1 WHERE candidateid = '$candidateid'" ;
                mysql_select_db('election2016');
                $retval = mysql_query( $sql, $conn );

                if(! $retval ) {
                   die('Could not update data: ' . mysql_error());
                }
                echo "Updated data successfully\n";

                mysql_close($conn);
             }
    ?>
<html>
    <center>
    <font size="2" face = "century gothic">
    <?php
    $con=mysqli_connect("localhost","root","","election2016");
    // Check connection
    if (mysqli_connect_errno())
    {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $result = mysqli_query($con,"SELECT * FROM candidate_info");
    ?>
    <form method = "post" action = "<?php $_PHP_SELF ?>">
    <?php
    echo "<table border='1'>
    <tr>
    <th>Candidate Name</th>
    <th>Position</th>
    <th>Vote</th>
    <th>Number of Votes</th>
    </tr>";
    while ($row = mysqli_fetch_array($result)) {
        echo "<tr>";
        echo "<td>" . $row['candidatename'] . "</td>";
        echo "<td>" . $row['position'] . "</td>";
        echo "<td><input type='radio' name='candidateid' value='".$row['candidateid']."' >";
        echo "<td>" . $row['numberofvotes'] . "</td>";
    }
    echo "</table>";
    mysqli_close($con);
    ?>

    <br>
    <br>
    <input name = "update" type = "submit" id = "update" value = "update">
    </form>
    </center>
    </font>
    </html>

Answer 3

需要注意的要点：

无需创建单独的<form></form>来发送candidateid以更新numberofvotes。
如果您在同一页面提交，请避免多个数据库连接。
您的数据库表candidate_info字段名称与您在<table><tr></tr></table>中写的内容不匹配。因此，请使用确切的列名称数据库表中的内容。
将完整的<table></table>放入<form></form>。
由于您正在寻找单个候选值以获得更新，因此radio按钮非常有用。如果需要更新多个候选值，则必须使用名称为array类型的复选框。

更新代码：

<html>
    <center>
        <font size="2" face = "century gothic">
        <?php
        $con=mysqli_connect("localhost","root","","election2016");
        if (mysqli_connect_errno()) {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
        ?>
        <form method="post" action = "<?php $_PHP_SELF ?>">
            <?php
            $result = mysqli_query($con,"SELECT * FROM candidate_info");
            echo "<table border='1'>
                <tr>
                    <th>Candidate Name</th>
                    <th>Position</th>
                    <th>Vote</th>
                    <th>Number of Votes</th>
                </tr>";
                while ($row = mysqli_fetch_array($result)) {
                    echo "<tr>";
                        echo "<td>" . $row['candidatename'] . "</td>";
                        echo "<td>" . $row['position'] . "</td>";
                        echo "<td><input type='radio' name='candidateid' value='".$row['candidateid']."' ></td>";
                        echo "<td>" . $row['numberofvotes'] . "</td>";
                        $candidateID=$row['candidateid'];
                    }
            echo "</table>";
            mysqli_close($con);
            ?>
            <br>
            <br>
            <input name = "update" type = "submit" id = "update" value = "update">
        </form>
    </center>
    </font>
</html>

<?php
if(isset($_POST['update'])) {

    $dbhost = 'localhost';
    $dbuser = 'root';
    $dbpass = '';
    $candidateid = $_POST['candidateid'];
    $conn = mysql_connect($dbhost, $dbuser, $dbpass);

    if(! $conn ) {
         die('Could not connect: ' . mysql_error());
    }

    $candidateid = $_POST['candidateid'];

    $sql = "UPDATE candidate_info SET numberofvotes = 1 WHERE candidateid = '$candidateid'" ;
    mysql_select_db('election2016');
    $retval = mysql_query( $sql, $conn );

    if(! $retval ) {
         die('Could not update data: ' . mysql_error());
    }
    echo "Updated data successfully\n";

    mysql_close($conn);
}
?>

在PHP中的未定义索引

3 个答案: