php - 我想使用带有 AJAX 的文本框搜索

Question

您好，我正在尝试弄清楚如何使用我的文本框通过 AJAX 进行实时搜索，然后使用数据库中的数据更新页面。但我希望它与以下标题一起输出:

            <th>Medewerker</th>
            <th>Datum overboeking</th>
            <th>Juliette geinformeerd</th>
            <th>Boekingsnummer</th>
            <th>Land</th>
            <th>Huiscode</th>
            <th>Aanbieder</th>
            <th>Contractvorm</th>
            <th>Huursom</th>
            <th>Periode van</th>
            <th>Periode tot</th>
            <th>Distributiekanaal</th>
            <th>Naam klant</th>
            <th>Reden</th>
            <th>Nieuwe boeking</th>
            <th>Reden geen nieuwe boeking</th>
            <th>Boete aanbieden</th>
            <th>Administratie geinformeerd voor inhouding boete</th>
            <th>Bedrag van boete</th>
            <th>Coulance Happyhome</th>
            <th>Opmerking</th>
            <th>Id</th>                

PHP

$pdo = new PDO('mysql:host=localhost;dbname=records', 'root', '***');
$select = 'SELECT *';
$from = ' FROM overboekingen';
$opts = isset($_POST['filterOpts'])? $_POST['filterOpts'] : array('');

 if (in_array("naam_klant", $opts)){
"naam_klant = '{$opts['naam_klant']}%' ";
 }  

 $sql = $select . $from . $where;
 $statement = $pdo->prepare($sql);
 $statement->execute();
 $results = $statement->fetchAll(PDO::FETCH_ASSOC);
 $json = json_encode($results);

 echo($json);

HTML

        <td>
            <input type="text" name="naam_klant" size="20" id="naam_klant" onkeyup="lookup(this.value);" onblur="fill();" >    
            <div class="suggestionsBox" id="suggestions" style="display: none;">
            <div class="suggestionList" id="autoSuggestionsList">
            </div>
            </div>
        </td>

AJAX

<script>
function makeTable(data) {
    var tbl_body = "";
    $.each(data, function() {
        var tbl_row = "";
        $.each(this, function(k , v) {
            tbl_row += "<td>"+v+"</td>";
        })
        tbl_body += "<tr>"+tbl_row+"</tr>";
    })
    return tbl_body;
}

function getEmployeeFilterOptions(){
    var opts = [];
    $checkboxes.each(function(){
        if(this.checked){
            opts.push(this.name);
        }
    });
    return opts;
}

function getEmployeeFilterOptions2(){
    var opts = [];
    $onchange.each(function(){
        if(this.checked){
            opts.push(this.name);
        }
    });

    return opts;
}

function updateEmployees(opts){
    $.ajax({
        type: "POST",
        url: "submit.php",
        dataType : 'json',
        cache: false,
        data: {filterOpts: opts},
        success: function(records){
            $('#employees tbody').html(makeTable(records));
        }
    });
}

var $checkboxes = $("input:checkbox");
$checkboxes.on("change", function(){
    var opts = getEmployeeFilterOptions();
    updateEmployees(opts);
});

updateEmployees();

</script>

但是我卡住了从这里开始做什么，代码在 atm 上不起作用。

Answer 1

我建议您首先修复返回结果的 php 脚本。

一个明显的错误是您在此 if 子句中的语句:

 if (in_array("naam_klant", $opts)){
"naam_klant = '{$opts['naam_klant']}%' ";
 }  

我认为你正在尝试设置 where 变量，所以它应该是这样的:

 if (in_array("naam_klant", $opts)){
$where = " WHERE naam_klant = '{$opts['naam_klant']}%' ";
 }