如何检索mysql表并输出到xml但使用URL中的变量

Question

该脚本可以从mysql表中提取数据，然后将其导出到xml中。

我如何通过将变量传递到URL并输出到XML来检索msgid

这应该很容易，但是我似乎做对了。

应该是这样的：

http：//localhost/test.php？msgid = d503dba6-44b8-4ba0-ae9a-8d6743a914ee

<data>
 <acceptreport>
  <id>4692</id>
  <username>test1</username>
  <msgid>d503dba6-44b8-4ba0-ae9a-8d6743a914ee</msgid>
  <sender>TOPS</sender>
  <receiver>523452345</receiver>
  <acceptedfordeliverytime>2016-06-21 09:04:16</acceptedfordeliverytime>
  <deliveredtohandsettime>2016-06-21 09:04:00</deliveredtohandsettime>
  <operator>AIRLINK</operator>
  <status>deliveredtohandset</status>
 </acceptreport>
</data>

这是脚本

<?php
//database configuration
$config['mysql_host'] = "192.168.1.1";
$config['mysql_user'] = "test";
$config['mysql_pass'] = "pass#word";
$config['db_name']    = "testdb";
$config['table_name'] = "box";

//connect to host
mysql_connect($config['mysql_host'],$config['mysql_user'],$config['mysql_pass']);
//select database
@mysql_select_db($config['db_name']) or die( "Unable to select database");

/**
 * @param mysql_resource - $queryResult - mysql query result
 * @param string - $rootElementName - root element name
 * @param string - $childElementName - child element name
 */
function sqlToXml($queryResult, $rootElementName, $childElementName)
{
    $xmlData = "<?xml version=\"1.0\" encoding=\"ISO-8859-1\" ?>\n";
    $xmlData .= "<" . $rootElementName . ">";

    while($record = mysql_fetch_object($queryResult))
    {
        /* Create the first child element */
        $xmlData .= "<" . $childElementName . ">";

        for ($i = 0; $i < mysql_num_fields($queryResult); $i++)
        {
            $fieldName = mysql_field_name($queryResult, $i);

            /* The child will take the name of the table column */
            $xmlData .= "<" . $fieldName . ">";

            /* We set empty columns with NULL, or you could set
                it to '0' or a blank. */
            if(!empty($record->$fieldName))
                $xmlData .= $record->$fieldName;
            else
                $xmlData .= "null";

            $xmlData .= "</" . $fieldName . ">";
        }
        $xmlData .= "</" . $childElementName . ">";
    }
    $xmlData .= "</" . $rootElementName . ">";

    return $xmlData;
}

/* Sql query */
$result = mysql_query("SELECT id, username, msgid, sender, receiver, acceptedfordeliverytime, deliveredtohandsettime, operator, status FROM outbox  WHERE username = 'top2' ORDER BY id DESC LIMIT 1");

/* If you want to process the returned xml rather than send it
    to the browser, remove the below line.
*/
header("Content-Type: application/xml");
echo sqlToXml($result, "data", "acceptreport");
?>

Answer 1

您传递了错误的变量以获取where子句中的数据。

$result = mysql_query("SELECT id, username, msgid, sender, receiver, acceptedfordeliverytime, deliveredtohandsettime, operator, status FROM outbox  WHERE username = 'burs2' ORDER BY id DESC LIMIT 1");

您需要在url中传递username以获取动态数据

$result = mysql_query("SELECT id, username, msgid, sender, receiver, acceptedfordeliverytime, deliveredtohandsettime, operator, status FROM outbox  WHERE username = '".$_REQUEST['username']."' ORDER BY id DESC LIMIT 1");

您不是通过msgid提取数据。 您应该呼叫的网址应为以下网址

http://localhost/test.php?username=test1