Question

我使用ajax在数据库中搜索值并填充下拉列表。这是我的ajax代码：

$output = '' ;
if (isset($_POST['searchVal'])){
$searchq = $_POST['searchVal'];
$st = $_POST['st'];
$query = mysqli_query($link, "SELECT DISTINCT title FROM `wp_locations` WHERE state='".$st."'  AND `title` LIKE  '%".$searchq."%' LIMIT 10")or     die("Could not search!");
if (!mysqli_query($link,$query))
{
        echo("Error description: " . mysqli_error($link));
}
$count = mysqli_num_rows($query);
if($count == 0){
    $output = '<option>No results!</option>';
}else{
while($row = mysqli_fetch_array($query)){
    $suburb = $row['title'];
    $postcode = $row['postcode'];

?>
    <option value="<?php echo $suburb; ?>"><?php echo $suburb; ?>    </option>
<?php
} // while
} // else
} // main if

在这里，我获取用户输入的区域并在数据库中搜索该区域中的郊区。这是我的PHP代码：

postcode

我想要做的是，将Suburb:* <input type="text" name="suburb" list="sbb" required size="30" value="<?php echo $suburb; ?>" onkeyup="searchq6()" id="output"> <datalist id="sbb" name="taskoption6" > <option> </option> </datalist> Postcode:* <input type="number" name="postcode" required value="<?php echo $postcode; ?>" id="postcode">值发送到另一个输入框。这就是我的html外观

$("#postcode").html(postcode);

只需添加public static void test(int x) { if (x > 1) { Console.WriteLine(x); } }即可。如何从php文件中发送值？

Answer 1

只需创建一个包含三个值的数组。 1.成功是真是假 2.如果成功则返回值的数据数组否则为null 3.成功价值为假的错误原因。

因此，您的一个错误结果将是：

$result = ['success' => false, 'data' => null, 'error' => 'error connecting db'];

您的成功结果将是：

$result = ['success'=> true, 'data'=> ['suburbs'=> $suburbs, 'pincode' => $pincode], 'error' => null];

现在将此结果数组发送为json。

echo json_encode($result);

然后解析comment by @MistahFiggins并填充你所在的reult。

Answer 2

在PHP文件中：

// Output is array, will be send as JSON data
$output = array("suburbs"=>'', "postcodes"=>array());

if (isset($_POST['searchVal'])){
    $searchq = $_POST['searchVal'];
    $st = $_POST['st'];
    $query = mysqli_query($link, "SELECT DISTINCT title FROM `wp_locations`     WHERE state='".$st."'  AND `title` LIKE  '%".$searchq."%'     LIMIT 10")or     die("Could not search!");
    if (!mysqli_query($link,$query)) {
        echo("Error description: " . mysqli_error($link));
    }
    $count = mysqli_num_rows($query);
    if($count == 0){
        $output["suburbs"] = '<option>No results!</option>';
    }else{
        while($row = mysqli_fetch_array($query)){
            $suburb = $row['title'];
            $postcode = $row['postcode'];
            $output["suburbs"] .= "<option value=\"$suburb\">$suburb  </option>";
            $output["postcodes"][$suburb] = $postcode;
        } // while
    } // else
    // Send result
    echo json_encode($output);
} // main if

在HTML的AJAX部分：

// Make global variable that holds POSTCODES:
var postcodes = {};
function searchq6(){
    var searchstate = $("input[name='region']").val();
    var searchTxt = $("input[name='suburb']").val();
    $.post("search-suburb.php", {searchVal: searchTxt, st:searchstate})
    .done(function(response) {
        var responseObj = jQuery.parseJSON(response);
        $("#sbb").html(responseObj["suburbs"]);
        postcodes = responseObj["postcodes"];
    }); 
}

在HTML中：

 Suburb:* <input type="text" onselect="change_post(this)" onchange="change_post(this)" name="suburb" list="sbb" required size="30"  value="<?php echo $suburb;  ?>" onkeyup="searchq6()" id="output">
    <datalist id="sbb" name="taskoption6" >
            <option> </option>
    </datalist>
 Postcode:* <input type="number" name="postcode" required value="<?php echo $postcode;  ?>"  id="postcode">
 <script>
      //var postcodes = { "1":"First", "2":"Second", "3":"Third", "4":"Fourth" };
        function change_post(out) {
            //console.log(postcodes[ out.value ]!=undefined);
            if ( postcodes[ out.value ]!=undefined )
                document.getElementById("postcode").innerHTML = postcodes[ out.value ];
        }
 </script>

如何通过ajax向php发送第二个值

2 个答案: