php - 在 PHP 中使用 ajax 预填充相应国家/地区的州详细信息

Question

选择任何国家/地区后，我需要在 frmState 下拉框中预填充州详细信息。但我是ajax新手，对它的了解还不够。你能帮我完成这个日常工作吗？下面给出。它可能值很多钱。提前致谢。

我在 mySql 中有两个表..

tbl_国家

columns: country_id(auto inc id) | country_name | country_code

tbl_状态

columns: state_id(auto inc id) | state_name | country_name

HTML代码

<select class="txtselectcomp" name="frmCountry" id="frmCountry" onChange="showHint(this.value);" onblur="checkEmptyData('frmCountry');" required="required" >
<option value="">Select  country</option>
    <?php
for ($i = 0; $i < count($strCountries); $i++) { ?>
         <option value="<?php
    echo $strCountries[$i]['country_name']; ?>" <?php
    if ($_POST['frmCountryName'] == $strCountries[$i]['country_name']) { ?> selected="selected" <?php
    } ?>><?php
    echo ucfirst($strCountries[$i]['country_name']); ?></option>
    <?php
} ?>
</select> 

<select class="txtselectcomp" name="frmState" id="frmState" onblur="checkEmptyData('frmState');" required="required" >
<option value="">Select State</option>
</select>   

Ajax 脚本

<script language="javascript">
function showHint(str)
{
    var xmlhttp;
    if (str.length==0) {
        document.getElementById("frmState").innerHTML="";
        return;
    }
    if (window.XMLHttpRequest) {
        xmlhttp=new XMLHttpRequest();
    }
    else {
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange=function() {
        if (xmlhttp.readyState == 4 && xmlhttp.status==200) {
            document.getElementById("frmState").innerHTML=xmlhttp.responseText;
        }
    }
    xmlhttp.open("GET","findState.php?value="+str,true);
    xmlhttp.send(null);
}
</script>

函数

$strCountries = doSelectRecordsById('tbl_country', 'country_id, country_name', 'country_id', ' order by country_name ASC', '');

function doSelectRecordsById($strTableName, $strSelect='*', $strTableId, $strOrderBy = NULL, $strFieldValue = NULL)
    {
        if ($strFieldValue != NULL) {
            $strWhereClause = " WHERE $strTableId = '".(int)$strFieldValue."'";
        }

        $strSqlRecords = "SELECT $strSelect FROM $strTableName $strOrderBy $strWhereClause";
        $strSqlData = SelectQry($strSqlRecords);

        if ($strWhereClause == '') {
            return $strSqlData;
        } else {
            return $strSqlData[0];
        }
    }

function doSelectState($strIdent)
    {
        $strSelectsSql = "Select * from tbl_states Where country_name='".$strIdent."' ";
        $strSelectsResult = SelectQry($strSelectsSql);
        return $strSelectsResult[0];
    }

findState.php

  <?php
    include ("config.php");
    $country_name = $_GET["value"];
    $sql = doSelectState($country_name);
    ?>
        <select name="State" id="State">
        <?php
    for ($i = 0; $i < count($sql); $i++) {
        $id = $sql[$i]['state_id'];
        $state = $sql[$i]['state_name'];
        echo '<option value="' . $id . '">' . $state . '</option>';
    }
   ?>
   </select>

Answer 1

<?php
include ("config.php");
$country_name = $_GET["value"];
$sql = doSelectState($country_name);
?>
    <select name="State" id="State">
    <?php
for ($i = 0; $i < count($sql); $i++) {
    $id = $sql[$i]['state_id'];
    $state = $sql[$i]['state_name'];
    echo '<option value="' . $id . '">' . $state . '</option>';
}

?>

在这里您不必创建选择元素。你可以这样做

<?php
include ("config.php");
$country_name = $_GET["value"];
$sql = doSelectState($country_name);
for ($i = 0; $i < count($sql); $i++) {
    $id = $sql[$i]['state_id'];
    $state = $sql[$i]['state_name'];
    echo '<option value="' . $id . '">' . $state . '</option>';
}

？>

这样，当将数据返回到浏览器时，您将获得选项字段，而不是选择，它可以是选择的内部 html。您现在要做的是将 SELECT 元素插入到选择元素中