根据用户搜索输入，从数据库中获取并存储id到不同的表

Question

我有多个动态文本框，用户输入车牌号码。我正在尝试从数据库中检索用户输入的牌照＃对应的id，但我无法得到它。我想将已经存在于数据库中的牌照＃的 id 存储到另一个表中，而不是将牌本本身存储，但它总是返回null。

提前致谢。

search.js

function firetruckAddingTextBoxes() {
var NumOfText = $("#NumOfTextBoxes2").val();
$('.NewlyCreatedSelectBoxes2').empty();
var txtBox = "";

for (i = 0; i < NumOfText; i++) {
    txtBox += '<input type="text" name="firetruck[]" class="search" placeholder="License Plate No" required/><br>';
}
$('.NewlyCreatedSelectBoxes2').append(txtBox);



$(".NewlyCreatedSelectBoxes2").on("keyup", ".search", function () {
    var search_term = $(this).val();

    $('#search_results2').html('Searching database...');

    if (search_term !== '') {
        $.post('php/firetruck_search.php', { search_term: search_term }, function (data) {
            $('#search_results2').html(data);
        });
    } else {
        $('#search_results2').html('Not Found');
    }

});
return false;

}

---

的search.php

<?php
require('connectdb.php'); 

if(isset($_POST['search_term']))
{
    $search_term = mysql_real_escape_string(htmlentities($_POST['search_term']));

    if(!empty($search_term))
    {
        $search = mysql_query("SELECT `fireTruckID`, `licensePlateNo` FROM `firetruckinfo` WHERE `licensePlateNo` LIKE '%$search_term%'");
        $result_count = mysql_num_rows($search);

        $suffix = ($result_count != 1) ? 's' : '';
        echo '<p>Your search for ', $search_term, ' returned ', $result_count, ' result', $suffix, '</p>';

        while($results_row = mysql_fetch_assoc($search))
        {
            echo '<p>', $results_row['licensePlateNo'], '</p>';
        }
    }
}
?>

---

occurrence.php

$firetrucks = $_POST['firetruck'];
        foreach($firetrucks as $firetruck):
            $vehicleID = mysql_query("SELECT `fireTruckID` FROM `firetruckinfo` WHERE `licensePlateNo`='$firetruck'");
            $row2 = mysql_fetch_assoc($vehicleID);

            $query3 = "INSERT INTO `truckonscene`(`FireTruckInfo_fireTruckID`, `IncidenceOfFire_incidentID`)
                   VALUES(
                   '".mysql_real_escape_string($row2['fireTruckID'])."',
                   '".mysql_real_escape_string($incidentid)."')";

            mysql_query($query3) or die(mysql_error());
        endforeach;