【发布时间】:2014-12-26 17:40:28
【问题描述】:
我有一个名为“customers”的数据库和五个名为 name、address、city、cap、country 的数组。在我的数据库中,我想创建 10 条记录,其中包含有关每个客户的信息。使用我的代码,所有数据都插入到数据库中,但首先插入了所有名称、所有地址等,所以我得到了 50 条记录而不是 10 条。我该如何解决?
<?php
$name = array('Alfreds Futterkiste','Ana Trujillo Emparedados','Antonio Moreno Taqueria','Around the Horn','Berglunds snabbkop','Blauer See Delikatessen','Blondel pere et fils ','Bolido Comidas preparadas','Bon app','Bottom-Dollar Marketse');
$address = array('Obere Str. 57','Avda. de la Constitucion 2222','Mataderos 2312','120 Hanover Sq.','Berguvsvagen 8','Forsterstr. 57','24, place Kleber','C/ Araquil, 67','12, rue des Bouchers','23 Tsawassen Blvd');
$city = array('Berlin','Mexico D.F.','Mexico D.F.','London','Lulea','Mannheim','Strasbourg','Madrid','Marseille','Tsawassen');
$cap = array('12209','05021','05023','WA1 1DP','S-958 22','68306','67000','28023','13008','T2F 8M4');
$country = array('Germany','Mexico','Mexico','UK','Sweden','Germany','France','Spain','France','Canada');
$conn= mysqli_connect('127.0.0.1','root','','preesame') or die("Connection failed: " . $conn->connect_error);
foreach ($name as $key=>$value)
{
$namess = mysqli_real_escape_string($conn,$value);
$addressess = mysqli_real_escape_string($conn,$addresses);
$citiess = mysqli_real_escape_string($conn,$cities);
$capss = mysqli_real_escape_string($conn,$caps);
$countriess = mysqli_real_escape_string($conn,$countries);
$insert = mysqli_query($conn,"INSERT INTO customers (id,name,address,city,cap,country) VALUES ('',$namess[$key],$addressess[$key],$citiess[$key],$capss[$key],$countriess[$key])");
}
$conn->close();
?>
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标签: php mysql database mysqli insert