语句导致while循环错误

时间:2013-05-31 17:15:55

标签: php

我编写了一些PHP / SQL来填充选择列表,其中包含来自数据库的数据,其中选项是根据文本框中的先前条目填充的。$ pk被完全接受并且是有效的ARTICLE_NO,并且查询有效完全由mysql直接执行。我在每个事件之后都放置了输出语句,并且在执行时除了tetsing之外的所有事永远不会输入while循环,我不确定为什么。这是我的代码:

编辑:我已经将问题缩小到返回0行的事实,但我不知道为什么因为phpmyadmin中的相同查询给出了正确的结果。

    if (!$getRecords->fetch()) {
    printf("<p>ErrorNumber: %d\n", $getRecords->errno);
}

它显示errno为0.因此没有提取记录,并且没有错误,但它是一个有效的查询。

<?php
ini_set('display_errors', '1');
error_reporting(E_ALL);
$pk = $_GET["pk"];
$con = mysqli_connect("localhost", "", "", "");
if (!$con) {
    echo "Can't connect to MySQL Server. Errorcode: %s\n". mysqli_connect_error();
    exit;
}
$con->set_charset("utf8");
echo "test outside loop";
if(1 < 2) {
    echo "test inside loop";
    $query1 = 'SELECT ARTICLE_NO FROM AUCTIONS WHERE ARTICLE_NO = ?';
    if ($getRecords = $con->prepare($query1)) {
echo "inside second loop";
        $getRecords->bind_param("i", $pk);
echo "test after bind param";
        $getRecords->execute();
echo "test after bind execute";
        $getRecords->bind_result($ARTICLE_NO);
echo "test after bind result";
        while ($getRecords->fetch()) {
            echo "test inside while";
            echo "<h1>".$ARTICLE_NO."</h1>";
        }
    }
}

编辑:

我尝试使用此代码:

<?php
$mysqli = new mysqli("localhost", "", "", "");
$pk = $_GET["pk"];
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}
/* prepare statement */
if ($stmt = $mysqli->prepare("SELECT ARTICLE_NAME, WATCH FROM AUCTIONS WHERE ARTICLE_NO = ? LIMIT 5")) {
    $stmt->bind_param("i", $pk);
    $stmt->execute();
    /* bind variables to prepared statement */
    $stmt->bind_result($col1, $col2);
    /* fetch values */
    while ($stmt->fetch()) {
        printf("%s %s\n", $col1, $col2);
    }
    /* close statement */
    $stmt->close();
}
/* close connection */
$mysqli->close();
?>

这没有$ pk,如果我拿走参数它工作正常。通过GET获取pk不是问题,因为如果我分配$ pk = 1;相反,它仍然失败。

编辑:问题是mysqli无法处理bigint,我现在使用k作为字符串,它工作正常。

2 个答案:

答案 0 :(得分:0)

我注意到的第一件事是数据库的名称。它应该是一个字符串'twa312',我还检查了$ _POST ['post']检查下面的代码:

    <?php

    $conn = mysql_connect("localhost", "*****", "*****");
    mysql_select_db('twa312', $conn)
    or die ('Database not found ' . mysql_error() );

    $options = '<option value="0"></option>';
    if(isset($_POST['post']))
    {
         $postcode = $_POST["post"];


    $sql = "SELECT school_info.Name AS Name, local_schools.postcodeID AS postcode FROM school_info INNER JOIN local_schools ON local_schools.schoolID = school_info.schoolID";
    $sql = $sql . " where postcode = '$postcode' ";

    $rs = mysql_query($sql, $conn)
    or die ('Problem with query' . mysql_error());

    while ($row=mysql_fetch_array($rs)) {
    $name=$row["Name"];
    $options .= '<OPTION VALUE="' . $name . '">' . $name ."</option>";
    }
}
else
{
$options = '';
}

    ?>

您也没有定义$ msgp变量。

希望它对您有所帮助,随时向我提出任何其他问题,并发布您的错误。

答案 1 :(得分:0)

这样的事情应该有效。您没有在查询中包含valid_postcodes表,因此服务器与不存在的字段混淆。

if (isset($postcode) && !empty($postcode)) {
$sql = "SELECT school_info.Name AS Name, local_schools.postcodeID AS postcode FROM school_info INNER JOIN local_schools ON local_schools.schoolID = school_info.schoolID";
$sql .= " JOIN valid_postcodes ON local_schools.postcodeID = valid_postcodes.id ";
$sql .= " where postcode = '$postcode' ";

$rs = mysql_query($sql, $conn)
or die ('Problem with query' . mysql_error());

while ($row=mysql_fetch_array($rs)) {
    $name=$row["Name"];
    $options .= '<OPTION VALUE="' . $name . '">' . $name ."</option>";
}
}

此外,通过确保查询仅在知道变量中存在值时才运行,可以减少出错的几率。并使用PHP的.=连接减少必须写出变量名的次数。

检查返回的数组,如下所示:

$rs = mysql_query($sql, $conn)
or die ('Problem with query' . mysql_error());

echo "<pre>";
print_r(mysql_fetch_assoc($rs));
echo "</pre>";

while ($row=mysql_fetch_array($rs)) {