Question

首先我这样做：

$zip_code_array = mysql_query("SELECT * FROM zip_code WHERE (lon BETWEEN '$lng_min_rnd'   AND '$lng_max_rnd') AND (lat BETWEEN '$lat_min_rnd' and '$lat_max_rnd')") or die (mysql_error());
while($zip_code_cells = mysql_fetch_array($zip_code_array))
{
$zip_codes_raw = $zip_code_cells['zip_code'];
$zip_codes_in_distance .= $zip_codes_raw.", ";
}

这可以工作并吐出像这样的EX：07110,07111,07112等的邮政编码：但后来我这样做了：

$user_list_array = mysql_query("SELECT * FROM member_search WHERE IN($zip_codes_in_distance) AND gender = '$gender_of_interest_session'");
while($user_list = mysql_fetch_array($user_list_array))
{
$id_from_array = $user_list['id'];
$username_from_array = $user_list['user_name'];
$defaultpic_from_array = $user_list['defaultpic'];
$city_from_array = $user_list['city'];

}

我在

上获得了Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

$user_list_array = mysql_query("SELECT * FROM member_search WHERE IN($zip_codes_in_distance) AND gender = '$gender_of_interest_session'");

我似乎无法找出查询它出现的zipcodes列表的正确方法。当我将邮政编码结果转换为字符串并将所有邮政编码用逗号分隔时，我确实认为这与某些事情有关。

Answer 1

您的sql语句有错误。 $zip_codes_in_distance的结尾为,，导致失败。

应该是这样的。

$zips = array();
$zip_code_array = mysql_query("SELECT * FROM zip_code WHERE (lon BETWEEN '$lng_min_rnd'   AND '$lng_max_rnd') AND (lat BETWEEN '$lat_min_rnd' and '$lat_max_rnd')") or die (mysql_error());
while($zip_code_cells = mysql_fetch_array($zip_code_array))
{
    $zip_codes_raw = $zip_code_cells['zip_code'];
    $zips[]= $zip_codes_raw;
}

$zip_codes_in_distance = "'". implode("','" , $zips). "'";

Answer 2

首先，检查查询是否成功：

<?php
$result = mysql_query("SELECT something FROM foo WHERE bar = 'foo'");

if( $result === false )
{ # Query not a succes :(
  echo 'We cannot execute the query, the error: '.mysql_error(); // Remove mysql_error() when you put the site online
}
else
{ # the query was a succes
  if( mysql_num_rows() <= 0 )
  { # If there are 0 or less rows
    echo 'We cannot find a result';
  }
  else
  { # You get 1 or more rows
    // Use fetch_assoc, because this is faster then fetch_array and you don't need a fetch_array
    while( $row = mysql_fetch_assoc($result) )
    {
      echo $row['something'];
    }
  }
}

一些资源：

Answer 3

修正了它：而不是像这样拆分数组：

$zip_codes_in_distance .= $zip_codes_raw.", ";

我是这样做的：

$zip_codes_in_distance .= "'".$zip_codes_raw."', ";

然后对于MySQL IN运算符我就像这样做了：

IN($zip_codes_in_distance '0')

我添加了零，没有，（逗号），因为数组在每个邮政编码的末尾添加一个逗号。当它这样做时，它在字符串的末尾留下了额外的逗号，从而产生了SQL错误。 0只是将字符串放在字符串的末尾，与搜索结果无关。

MySQL从数组中搜索

3 个答案: