Question

我正在尝试使用php将mysql数据库中的项目传递到xml文件中。我有php代码创建xml文件。但是传递给它的值不是来自mysql数据库的值。该数据库有13个cols，有62行。我的foreach语句中有五个项目，当它们显示在Web屏幕上时，值输出如下：

找到的物业数量：62

1 1 1 1 1 1 1 1 1 1 1 1 1 1 R R R R R E E E E W W W W A A A A A h h h h h 1 1 1 1 1＆lt; ＆LT; ＆LT; ＆LT; ＆LT; 2 2 2 2 2 1 1 1 1 1 1 1 1 1

上面一行中有65个项目，这是我的foreach语句中的5个项目乘以数据库中的13个cols。我认为这与它有关。

以下是我的代码：

<?php

@$db = new mysqli('localhost', 'root', '', 'siamsatire');

if (mysqli_connect_errno()) {
    echo 'error connecting to db';
    exit;
}
$query = "SELECT * from events";
$result = $db->query($query);
$num_results = $result->num_rows;
echo 'Number of properties found : <strong>' . $num_results . '</strong><br><br>';

for ($i=0; $i < $num_results; $i++) { 

    $row = $result->fetch_object(); 

    $name = $row->name; 
    $subtitle = $row->sub_title; 
    $date = $row->display_date; 
    $description = $row->slug; 
    $photo= $row->photo; 
    $thumb= $row->thumb; 

    /*echo '<tr>';
    echo "<td>$name</td>";
    echo "<td>$subtitle</td>";
    echo "<td>$date</td>";
    echo "<td>$description</td>";
    echo "<td>$photo</td>";
    echo "<td>$thumb</td>";1+0
    echo '<tr>';*/

} 

$doc = new DOMDocument("1.0");
$doc->formatOutput = true;

$r = $doc->createElement("events");
$doc->appendChild( $r );

foreach($row as $fieldvalue)
{
    $b = $doc->createElement( "event" );

    $name1 = $doc->createElement( "title" );
    $name1->appendChild( $doc->createTextNode( $fieldvalue['title'] ));
    $b->appendChild( $name1 );

    $subtitle1 = $doc->createElement( "subtitle" );
    $subtitle1->appendChild($doc->createTextNode( $fieldvalue['subtitle'] ));
    $b->appendChild( $subtitle1 );

    $date1 = $doc->createElement( "display_date" );
    $date1->appendChild($doc->createTextNode( $fieldvalue['display_date'] ));
    $b->appendChild( $date1 );

    $description1 = $doc->createElement( "slug" );
    $description1->appendChild( $doc->createTextNode( $fieldvalue['slug'] ));
    $b->appendChild( $description1 );

    $photo1 = $doc->createElement( "photo" );
    $photo1->appendChild( $doc->createTextNode( $fieldvalue['photo'] ) );
    $b->appendChild( $photo1 );

    $thumb1 = $doc->createElement( "thumb" );
    $thumb1->appendChild( $doc->createTextNode( $fieldvalue['thumb'] ) );
    $b->appendChild( $thumb1 );

    $r->appendChild( $b );
}

echo $doc->saveXML();
$doc->save("write.xml");

$result->free();
$db->close();
?>

有没有人对我做错了什么有任何想法？

更新

@starx - 我根据您的代码更改了我的代码，这就是现在的样子。

<?php

    @$db = new mysqli( 'localhost', 'root', '', 'siamsatire');

    if (mysqli_connect_errno()) {
    echo 'error connecting to db';
    exit;
    }

    $query = "SELECT * from events";

    $result = mysql_query($query);  

    if(mysql_num_rows($result)) {
    $doc = new DOMDocument("1.0");
    $doc->formatOutput = true;

    while($row = mysql_fetch_assoc($result)) {
        $r = $doc->createElement( "events" );
        foreach($row as $field=>$value) {
            $tChild = $doc->createElement( $field );
            $tChild->appendChild( $doc->createTextNode($value) );
            $r->appendChild( $tChild );     
        }
        $doc->appendChild($r);
    }
        $doc->appendChild( $r );
        echo $doc->saveXML();
        $doc->save("write.xml");
    }

    //$result->free();
        //$db->close();
    ?>

这些是我得到的错误。'

Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\siamsatire1.php on line 12

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\xampp\htdocs\siamsatire1.php on line 12

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\siamsatire1.php on line 14'

你知道我为什么得到它们吗？

然后我将mysql_query更改为mysqli_query，将错误减少到：

Warning: mysqli_query() expects at least 2 parameters, 1 given in C:\xampp\htdocs\siamsatire1.php on line 12

Warning: mysql_num_rows() expects parameter 1 to be resource, null given in C:\xampp\htdocs\siamsatire1.php on line 14

Answer 1

这是一个更好，更正确的解决方案

$query = "SELECT * from events";
$result = mysql_query($query);
if(mysql_num_rows($result)) {
    $doc = new DOMDocument("1.0");
    $doc->formatOutput = true;

    while($row = mysql_fetch_assoc($result)) {
        $r = $doc->createElement( "events" );
        foreach($row as $field=>$value) {
            $tChild = $doc->createElement( $field );
            $tChild->appendChild( $doc->createTextNode($value) );
            $r->appendChild( $tChild );     
        }
        $doc->appendChild($r);
    }
    $doc->appendChild( $r );
    echo $doc->saveXML();
    $doc->save("write.xml");
}

如果需要，您可以将上述代码与您的库集成。

更新（问题更新后）

以下是使用mysqli

的工作解决方案

<?
@$db = new mysqli( 'localhost', 'root', '', 'siamsatire');
if (mysqli_connect_errno()) {
    echo 'error connecting to db';
    exit;
}
$query = "SELECT * from events";
$result = mysqli_query($db,$query);  
if(mysqli_num_rows($result)) {
    $doc = new DOMDocument("1.0");
    $doc->formatOutput = true;

        while($row = mysqli_fetch_assoc($result)) {
            $r = $doc->createElement( "events" );
            foreach($row as $field=>$value) {
                $tChild = $doc->createElement( $field );
                $tChild->appendChild( $doc->createTextNode($value) );
                $r->appendChild( $tChild );     
            }
            $doc->appendChild($r);
        }
        $doc->appendChild( $r );
        echo $doc->saveXML();
        $doc->save("write.xml");
}

//$result->free();
//$db->close();
?>

使用php将数据库记录更改为xml文件

更新

1 个答案:

更新（问题更新后）