Question

如何从列表菜单中排除某些国家/地区？现在我的代码将列出数据库中的所有国家/地区名称。

示例我想从列表菜单中排除阿尔巴尼亚国家/地区。如何根据这些代码实现它。

代码

    <?php $select_country=mysql_query("SELECT * FROM tbl_country ORDER BY country_name ASC")?>
    <select name="country"  onchange="return get_country(this.value);">
        <option value=""><?=SELECT_COUNTRY?></option>
        <? while($country=mysql_fetch_array($select_country)) {?>
           <option  <? if($_SESSION['getcountry']==$country['id']){ ?> selected="selected"<? }?> value="<?=$country['id']?>">
           <?=$country['country_name']?></option>
        <? } ?>
    </select>

Mysql的

id    country_name
1   Afghanistan
2   Aland Islands
3   Albania
4   Algeria
5   American Samoa
6   Andorra

Answer 1

从结果集中排除该行：

SELECT * FROM tbl_country WHERE country_name != "Albania" ORDER BY country_name ASC

或者你用PHP跳过它：

<?php
    while($country=mysql_fetch_array($select_country)) {
        if ($country['country_name'] == 'Albania') {
            continue;
        } ?>
    <option <?php if($_SESSION['getcountry']==$country['id']){ ?> selected="selected"<? }?> value="<?=$country['id']?>"><?=$country['country_name']?></option>
<?php } ?>

Answer 2

"SELECT * FROM tbl_country WHERE country_name <> "Albania" ORDER BY country_name ASC"

“＆lt;＆gt;”相当于“！=”并且它们都意味着“不等于”，理论上应该从表中选择country_name不等于定义的“阿尔巴尼亚”的所有记录。

我认为以下作品也是：

"SELECT * FROM tbl_country WHERE country_name NOT IN("Albania","other_country_name", "another_country_name") ORDER BY country_name ASC"

Answer 3

如果我理解了你的问题：

 mysql_query("SELECT * FROM tbl_country WHERE country_name <> 'Algeria' ORDER BY country_name ASC")

Answer 4

您可以设置要排除的国家/地区的数组...

<?php

$excludeArray = array('Albania','United States') ;

$select_country=mysql_query("SELECT * FROM tbl_country ORDER BY country_name ASC");

echo <<<HTML
    <select name="country"  onchange="return get_country(this.value);">
    <option value=""><?=SELECT_COUNTRY?></option>
HTML;

while($country=mysql_fetch_array($select_country)) {
  if (!in_array($country['country_name'],$excludeArray)) {
    $selected = ($_SESSION['getcountry']==$country['id'] ? 'selected="selected"' : '');
    echo "<option value='" . $country['id'] . "' " . $selected . ">" . $country['country_name'] . "</option>" ;
  }
echo "</select>";


?>

Answer 5

<?php $select_country=mysql_query("SELECT * FROM tbl_country ORDER BY country_name ASC")?>
    <select name="country"  onchange="return get_country(this.value);">
    <option value=""><?=SELECT_COUNTRY?></option>
    <?php 
       $exclude = array('Afghanistan', 'Andorra');
       while($country=mysql_fetch_array($select_country)) {
         if(!in_array($country['name'], $exclude){
    ?>
    <option  <? if($_SESSION['getcountry']==$country['id']){ ?> selected="selected"<? }?> value="<?=$country['id']?>"><?=$country['country_name']?></option>
    <? }} ?>
    </select>

Answer 6

为什么不在数据库上创建状态呢？例如 albania status show ，您可以使用SELECT FROM country WHERE status='show'

Php - 如何从mysql中排除数据

6 个答案: