Question

我正在尝试将xml数据插入到我的sql中，但它没有插入。如何为这种类型的xml编写foreach（xml-＆gt;）。在这个结构中动态生成xml。这是这种格式的示例xml

<?php
$xmlData =<<< END
<?xml version="1.0"?>
<Customer>
<id>1</id>
<name>Oluwafemi</name>
<address>Cresent Drive, TX</address>
<list>
<contact>56689</contact>
<telephone>5889745</telephone>
</list>
<offer>congrats</offer>
</Customer>
END;

$xml = simplexml_load_string($xmlData) or die("ERROR: Cannot create SimpleXML object");
$connection = mysqli_connect("localhost", "root", "", "Customers") or die ("ERROR: Cannot connect");

foreach ($xml->Customer as $Customer) {
$id = $Customer->id;
echo "$id";
$name =  $Customer->name;
$address = $Customer->address;

$sql = "INSERT INTO customerdata (id, name, address ) VALUES ('$id', '$name', '$address')";
mysqli_query($connection, $sql) or die ("ERROR: " .mysqli_error($connection) . " (query was $sql)");
}
mysqli_close($connection);
?>

Answer 1

假设您有多个客户......

$sXmlString =<<< END
<?xml version="1.0"?>
<Content>
<Customer>
<id>1</id>
<name>Oluwafemi</name>
<address>Cresent Drive, TX</address>
</Customer>
<Customer>
<id>2</id>
<name>Oluwafemi2</name>
<address>Cresent Drive 2, TX</address>
</Customer>
</Content>
END;

$oXml       = simplexml_load_string($sXmlString);
$sJson      = json_encode( $oXml );
$aContent   = json_decode( $sJson, TRUE );
$aCustomers = $aContent[ 'Customer' ];
var_dump( $aCustomers );
$iCountCustomers = count( $aCustomers );
for( $i = 0; $i < $iCountCustomers; ++$i )
{
    $sId      = $aCustomers[ $i ][ 'id' ];
    $sName    = $aCustomers[ $i ][ 'name' ];
    $sAddress = $aCustomers[ $i ][ 'address' ];
    var_dump( $sId );
    var_dump( $sName );
    var_dump( $sAddress );
}

Answer 2

您需要编辑XML文档，使用以下标记封装文档：
<Customers> ... </Customers>

因此，代码的开头部分应如下所示：

<?xml version="1.0"?>
$sXmlString =<<< END
<Customers>
  <Customer>
    <id>1</id>
    <name>Oluwafemi</name>
    <address>Cresent Drive, TX</address>
  </Customer>
</Customers>
END;

编辑：OP无法编辑提供的XML，我认为它的结构是新客户将添加其他ID，名称，地址实体。以下是OP的更新代码：

<?php
$xmlData =<<< END
<?xml version="1.0"?>
<Customer>
  <id>1</id>
  <name>Oluwafemi</name>
  <address>Cresent Drive, TX</address>
  <id>2</id>
  <name>Rob</name>
  <address> 123 Longhorn </address>
</Customer>
END;

$xml = simplexml_load_string($xmlData) or die("ERROR: Cannot create SimpleXML object");
$connection = mysqli_connect("localhost", "root", "", "Customers") or die ("ERROR: Cannot connect");

/* Assumes that the number of IDs = number of customers */
$size = sizeOf($xml->id);
$i = 0; //index

/* Add each customer to the database, See how we reference it as    $xml->ENTITY[INDEX] */
while($i != $size) 
{
    //echo $xml->id[$i]; //Test
    $sql = "INSERT INTO customerdata (id, name, address ) VALUES ('$xml->id[$i]', '$xml->name[$i]', '$xml->address[$i]')";
    mysqli_query($connection, $sql) or die ("ERROR: " .mysqli_error($connection) . " (query was $sql)");

    $i++; //increment index
}

mysqli_close($connection);

我还在其他客户ID中添加了数据。你可以删除它，代码仍然可以正常工作。

使用PHP将XML数据插入MySQL表

2 个答案: