PHP和MySQLi查詢始終返回0

Question

以下代碼似乎無法正常工作。 我是PHP和jQuery的新手。

的PHP：

<?php

//if (!defined('BOOTSTRAP')) { die('Access denied'); }

//if we got something through $_POST
if (isset($_POST['postcode_locator_search'])) {
    // here you would normally include some database connection
    //include('config.local.php');

    //Open a new connection to the MySQL server
    $mysqli = new mysqli('localhost','test','c@W)ukmd[0bm','test');

    //Output any connection error
    if ($mysqli->connect_error) {
        die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
    }

    // never trust what user wrote! We must ALWAYS sanitize user input
    $postcode_q = mysqli_real_escape_string($mysqli, $_POST['postcode_locator_search']);
    $postcode_q = htmlentities($postcode_q);

    // A select query. $result will be a `mysqli_result` object if successful
    $result = mysqli_query("SELECT description FROM cscart_postcode_location_descriptions WHERE cscart_postcode_location_descriptions LIKE '%" . $postcode_q . "%' ORDER BY cscart_postcode_location_descriptions LIMIT 1");

    if($result === false) {
        // Handle failure - log the error, notify administrator, etc.
        echo '1';
    } else {
        // Fetch all the rows in an array
        echo '0';
    }

    $mysqli->close();

}
?>

JS / HTML：

{assign var="prod_id" value=$product.product_id}

<form action="search_postcode.php" method="post" class="postcode_locator_form" name="postcode_locator_form">
    <div class="ty-control-group">
        <label for="postcode_locator_search{$block.block_id}" class="ty-control-group__title">{__("postcode_search")}</label>
        <p class="filling-notice">{__("postcode_search_desc")}</p>
        <div class="ty-input-append ty-m-none">
            <input type="text" size="20" class="ty-input-text" id="postcode_locator_search" name="postcode_locator_search" value="{$postcode_locator_search.q}" />
            {include file="buttons/go.tpl" but_name="postcode_locator.search" alt=__("search")}
        </div>

    </div>
</form>

<div class="filling-status filling-success">
    <h3>Add filling to your bean bag</h3>
    <p>Searched postcode: <span class="searched-postcode"></span></p>
    <p class="beans-msg">{__("add_some_beans_success")} <a href="{"checkout.add_bean_bag_filling&product_id=`$product.product_id`"|fn_url}">{__("click_here")}</a></p>
</div>
<div class="filling-status filling-failure">
    <h3>Add filling to your bean bag</h3>
    <p>Searched postcode: <span class="searched-postcode"></span></p>
    <p class="beans-msg">{__("add_some_beans_error")}</p>
</div>

<script>
$(function() {

    $(".filling-status").hide();
    $(".postcode_locator_form .ty-btn-go").click(function() {
        // getting the value that user typed
        var searchString    = $("#postcode_locator_search").val();
        // forming the queryString
        var data            = 'postcode_locator_search='+ searchString;

        // if searchString is not empty
        if(searchString) {
            // ajax call
            $.ajax({
                type: "POST",
                url: "search_postcode.php",
                data: data,
                beforeSend: function(html) { // this happens before actual call
                    $(".searched-postcode").html(searchString);
                },
                success: function(data){ // this happens after we get results
                    console.log(data);
                    if(data == '0'){
                        $(".filling-status.filling-success").show();
                    } else  if(data == '1'){
                        $(".filling-status.filling-failure").show();
                    }
                }
            });    
        }
        return false;
    });
});
</script>

通訊正常，但是無論我搜索什么，它總是返回0作為成功，並且似乎不檢查數據庫的結果。

我需要的是如果我搜索某項並且是一個匹配項，則返回0作為成功，但是如果找不到，則返回一個匹配項作為失敗。

Answer 1

如果要檢索數據：

$result = mysqli_query("SELECT description FROMcscart_postcode_location_descriptions WHERE cscart_postcode_location_descriptions LIKE '%" . $postcode_q . "%' ORDER BY cscart_postcode_location_descriptions LIMIT 1");

if($result === false) {
    // Handle failure - log the error, notify administrator, etc.
    echo '1';
} else {
    // Fetch all the rows in an array
    while($row = mysqli_fetch_assoc($result)){
    echo $row['id']; //prints the resulted id
    }
}

Answer 2

使用mysqli_num_rows檢測是否有結果

if($result === false or mysqli_num_rows($result) === 0) {
    echo '1';
}

我建議將這種情況分成兩個if條件，以便您將錯誤與查詢分開處理而沒有結果