Question

我的php脚本：

$cont = array();
$slq_get_previous_country_list = "SELECT country FROM aw_countries_v2 WHERE id ='11120'";
$result_previous_country_list = mysql_query($slq_get_previous_country_list);
while($row = mysql_fetch_assoc($result_previous_country_list))
{
   $cont[] = $row['country'];
}

print_r($cont);                 
$not_data = array();    
$new_contry = array("Greece",
                    "Israel",
                    "Macedonia - The Frm Yugoslav Rep Of",
                    "Malta",
                    "India");


foreach($cont as $c)
{
  if(in_array($c, $new_contry)) 
    {
        echo $c."\n";
    }
    else
    {
        $not_data[] = $c;
    }   
}

print_r($not_data);

$cont数组：

Array
(
    [0] =>  Belgium
    [1] =>  Cyprus
    [2] =>  Czech Republic
    [3] =>  Fiji
    [4] =>  Finland
    [5] =>  Greece
    [6] =>  Iceland
    [7] =>  Ireland
    [8] =>  Israel
    [9] =>  Macedonia - The Frm Yugoslav Rep Of
    [10] =>  Malaysia
    [11] =>  Malta
    [12] =>  Monaco
    [13] =>  Poland
    [14] =>  South Africa
    [15] => Austria
)

$not_data数组：

Array
    (
        [0] =>  Belgium
        [1] =>  Cyprus
        [2] =>  Czech Republic
        [3] =>  Fiji
        [4] =>  Finland
        [5] =>  Iceland
        [6] =>  Ireland
        [7] =>  Malaysia
        [8] =>  Monaco
        [9] =>  Poland
        [10] =>  South Africa
        [11] => Austria
    )

结果：

array(
      [0] => austria
     );

我做错了什么？

Answer 1

你的字符串中有空格，in_array函数是空格和区分大小写的，所以尝试将其添加为前体：

while($row = mysql_fetch_assoc($result_previous_country_list))
{
   $cont[] = trim($row['country']);
}

您可能还想尝试strtolower来比较两个数组的字符串值，而不必担心。

此外，建议不要使用MySQL_，而是要开发 MySQLi ，甚至 PDO 。

in_array似乎不适用于数组

1 个答案: