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当我使用多项选择下拉列表从国家/地区表中选择国家/地区时，如何从国家/地区表中获取国家/地区列表？这是我的编码。

mysql表

CREATE TABLE `countries` (
  `countryID` varchar(3) NOT NULL default '',
  `countryName` varchar(52) NOT NULL default '',
  `localName` varchar(45) NOT NULL,
  `webCode` varchar(2) NOT NULL,
  `region` varchar(26) NOT NULL,
  `continent` enum('Asia','Europe','North America','Africa','Oceania','Antarctica','South America') NOT NULL,
  `latitude` double NOT NULL default '0',
  `longitude` double NOT NULL default '0',
  `surfaceArea` float(10,2) NOT NULL default '0.00',
  `population` int(11) NOT NULL default '0',
  PRIMARY KEY  (`countryID`),
  UNIQUE KEY `webCode` (`webCode`),
  UNIQUE KEY `countryName` (`countryName`),
  KEY `region` (`region`),
  KEY `continent` (`continent`),
  KEY `surfaceArea` (`surfaceArea`),
  KEY `population` (`population`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

表结构states：

CREATE TABLE `states` (
  `stateID` smallint(5) unsigned NOT NULL auto_increment,
  `stateName` varchar(50) NOT NULL default '',
  `countryID` varchar(3) NOT NULL,
  `latitude` double NOT NULL default '0',
  `longitude` double NOT NULL default '0',
  PRIMARY KEY  (`stateID`),
  KEY `stateName` (`stateName`),
  KEY `countryID` (`countryID`),
  KEY `unq` (`countryID`,`stateName`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

表的数据countries(LIMIT 0,500)

INSERT INTO `countries` (`countryID`, `countryName`, `localName`, `webCode`, `region`, `continent`, `latitude`, `longitude`, `surfaceArea`, `population`) VALUES 
  ('BRA','Brazil','Brasil','BR','South America','South America',-10,-55,8547403.00,170115000),
  ('CHN','China','Zhongquo','CN','Eastern Asia','Asia',35,105,9572900.00,1277558000),
  ('FRA','France','France','FR','Western Europe','Europe',47,2,551500.00,59225700),
  ('IND','India','Bharat/India','IN','Southern and Central Asia','Asia',28.47,77.03,3287263.00,1013662000),
   ('USA','USA','United States','US','North America','North America',38,-97,9363520.00,278357000);
COMMIT;

表的数据states(LIMIT 0,500)

INSERT INTO `states` (`stateID`, `stateName`, `countryID`, `latitude`, `longitude`) VALUES 
  (5,'California','USA',37.42,-122.06),
  (6,'Beijing','CHN',39.93,116.39),
  (9,'Iowa','USA',43.03,-96.09),
  (10,'New York','USA',40.76,-73.97),
  (12,'... ....','CHN',32.06,118.78);
COMMIT;

国家下拉菜单

<? $Type_sql="SELECT countryName FROM countries ORDER by countryName ASC";
                                $Type_result=mysql_query($Type_sql);
                                while($Type_rows=mysql_fetch_array($Type_result)){
                                echo "<option value='"; 
                                echo $Type_rows['countryName']; 
                                echo "'>";
                                echo $Type_rows['countryName'];
                                echo "</option>";
                                } ?>
            </select>

状态下拉列表

<? $Type_sql="SELECT stateName FROM countries c, states s where c.countryID = s.countryID ORDER by stateName ASC";
                                $Type_result=mysql_query($Type_sql);
                                while($Type_rows=mysql_fetch_array($Type_result)){
                                echo "<option value='"; 
                                echo $Type_rows['stateName']; 
                                echo "'>";
                                echo $Type_rows['stateName'];
                                echo "</option>";
                                } ?>
            </select>
4

3 回答 3

0

这可以通过将状态下拉列表的代码放入单独的 php 处理程序中来完成，稍作修改：

<select name="state">
<?php 
    $countryID= mysql_real_escape_string($_GET['countryID']);
    $Type_sql="SELECT stateName FROM states s where s.countryID = '$countryID' ORDER by stateName ASC";
                $Type_result=mysql_query($Type_sql);
                while($Type_rows=mysql_fetch_array($Type_result)){
                    echo "<option value='";
                    echo $Type_rows['stateName'];
                    echo "'>";
                    echo $Type_rows['stateName'];
                    echo "</option>";
                } 
?>
</select>

每当国家/地区选择列表中有 onChange 事件时，此脚本将由 javascript 使用 Ajax 调用。请参考 ajax 文档，例如http://www.javascriptkit.com/dhtmltutors/ajaxgetpost.shtml或使用 jQuery： http: //api.jquery.com/jQuery.get/

现在也可以像您的计划一样修改状态选择的内容，但是您需要两个选择列表，一个隐藏列表来存储您删除的条目或全部。

于 2013-03-12T08:57:00.417 回答
0

这是一个完整的工作代码。只需将此代码放在 php 页面中，并将 onChange 参数中的表单名称更改为您的页面名称。此外，更改mysql_select_db中的数据库名称。

我的页面名称是form1.php

<html>
<body>
<form name="form1">
<?php

if(isset($_REQUEST['stateId']))
{
    $stateId = $_REQUEST['stateId'];
}
else
{
    $stateId = "";
}
$conn = mysql_connect("localhost","root","");
$dbSelect = mysql_select_db("test");
$Type_sql="SELECT countryID,countryName FROM countries ORDER by countryName ASC";
                                $Type_result=mysql_query($Type_sql);
                                echo"Countries:<select onChange=\"window.location='form1.php?stateId='+this.value\">";
                                while($Type_rows=mysql_fetch_array($Type_result)){
                                    if($Type_rows['countryID']==$_REQUEST['stateId'])
                                    {
                                        echo "<option value='".$Type_rows['countryID']."' selected>";
                                        echo $Type_rows['countryName'];
                                        echo "</option>";
                                    }
                                    else
                                    {
                                        echo "<option value='".$Type_rows['countryID']."'>";
                                        echo $Type_rows['countryName'];
                                        echo "</option>";
                                    }
                                } 
                                echo"</select>";

 $Type_sql="SELECT stateName FROM  states where countryID = '$stateId' ORDER by stateName ASC";
                                $Type_result=mysql_query($Type_sql);
                    echo"States:<select><option value=''>--select--</option>";
                                while($Type_rows=mysql_fetch_array($Type_result)){
                                echo "<option value='"; 
                                echo $Type_rows['stateName']; 
                                echo "'>";
                                echo $Type_rows['stateName'];
                                echo "</option>";
                                } 
                    echo"</select>";

?>
</form>
</body>
</html>
于 2013-03-12T11:59:23.300 回答
0

嗨艾哈迈德，

请看下面的代码。我重新编写了代码并修改了多项选择 无论如何我都忽略了javascript验证。请继续努力，这个页面的名称是form3.php

<html>
<head>
<script type="text/javascript">
    function viewstates()
    { 
        jQuery('#langucountry').change(function () { });
    }
    $(function() {
        $('#basicmultiselect').multiSelect({select_all_min: 3});
        $('#langucountry').multiSelect({
            select_all_min: 3, no_selection: "Please select!", selected_text: " clicked" });
        $('#methods').multiSelect(); });
</script>
</head>
<body>
<form name="form3" action="form3.php" method="post">
<?php
/****** connection string *******/
/* start */
    $conn = mysql_connect("localhost","root","");
    $dbSelect = mysql_select_db("test");
/* end */

/*----checks whether form is submited -------*/
if(isset($_POST['cbocountry']))
{
    $lstStates = array();
        foreach($_POST['cbocountry'] as $countryCode)
        {
            $fetch_states="SELECT stateName FROM  states where countryID = '$countryCode' ORDER by stateName ASC";
            $Type_result=mysql_query($fetch_states);
            while($Type_rows=mysql_fetch_array($Type_result)){

                        if(!in_array($Type_rows['stateName'],$lstStates))
                            array_push($lstStates,$Type_rows['stateName']);
             } 
        }

}


                                $Type_sql="SELECT countryID,countryName FROM countries ORDER by countryName ASC";
                                $Type_result=mysql_query($Type_sql);
                                echo"Countries:<select  id=\"langucountry\" name=\"cbocountry[]\" class=\"arc90_multiselect\" multiple=\"multiple\" title=\"Languages\"\>";
                                while($Type_rows=mysql_fetch_array($Type_result)){

                                        echo "<option value='".$Type_rows['countryID']."'>";
                                        echo $Type_rows['countryName'];
                                        echo "</option>";
                                } 
                                echo"</select>";





                    echo"States:<select><option value=''>--select--</option>";
                            if(count($lstStates) > 0)
                            {
                                foreach($lstStates as $item)
                                {
                                    echo "<option value='"; 
                                    echo $item; 
                                    echo "'>";
                                    echo $item;
                                    echo "</option>";
                                } 
                            }
                    echo"</select>";

                    echo"<input type=\"submit\" value=\"submit\" text=\"submit\"/>";

?>
</form>
</body>
</html>

享受！！！！

于 2013-03-13T06:05:29.257 回答