Question

我有一个适用于城市搜索字段的jquery脚本脚本

<script type="text/javascript">   
  $(document).ready(function()
  { 
     var ac_config = 
     { 
        source: "demo_cities.php", 
        select: function(event, ui)
        { 
           $("#City").val(ui.item.City); 
           $("#Country").val(ui.item.Country); 
           $("#DestinationId").val(ui.item.DestinationId); 
        }, 
        minLength:3 
     }; 
     $("#City" ).autocomplete(ac_config); 
  }); 
</script>

和php代码是

<?php
   $cities= array(
      array('City'=>'Barcelona',
             Country=>'SP', 
             DestinationId=>'10001'),
             ....... );
   $term = trim(strip_tags($_GET['term']));   
   $matches = array(); 
   foreach($cities as $City)
   { 
      if(stripos($City['City'], $term) !== false)
      { 
         $City['value'] = $City['City']; 
         $City['label'] = "{$City['City']}";
         $matches[] = $City; 
      }
   }   
   $matches = array_slice($matches, 1, 7); 
   print json_encode($matches);
?>

如何让脚本仅按城市名称中的前3个字符进行搜索

Answer 1

在您的PHP代码中，您正在寻找一个包含城市名称中的字符串的城市。如果您想要显示以特定字符开头的城市，您可以使用substr

<?php
  $cities= array(
     array('City'=>'Barcelona',
         'Country'=>'SP', 
         'DestinationId'=>'10001'
     ),
     //more cities ... 
  );
  $term = strtolower(trim(strip_tags($_GET['term'])));   
  $matches = array(); 
  foreach($cities as $City)
  { 
     if(strtolower(substr($city['city'], 0, strlen($term))) == $term)
     { 
        $City['value'] = $City['City']; 
        $City['label'] = "{$City['City']}";
        $matches[] = $City; 
     }
  }   
  $matches = array_slice($matches, 1, 7); 
  print json_encode($matches);
?>

Answer 2

通常城市不会更改其名称，所以我将城市名称放在js文件中，然后让FE进行搜索，和自动完成部分似乎毫无疑问。

城市搜索jquery脚本与json列表

2 个答案: