Question

我找到了一个非常好的搜索查询here，但是此搜索查询使用了mysql database，现在我正在尝试将此代码调整为SQLite3 database但是使用我当前的代码我得到了这个错误

致命错误：在第79行的C：\ xampp \ htdocs \ xport \ searchtext.php中的非对象上调用成员函数fetchArray（）

以下是我当前的代码

<?php
require_once ("db.php");
$db = new MyDB();

$with_any_one_of = "";
$with_the_exact_of = "";
$without = "";
$starts_with = "";
$search_in = "";
$advance_search_submit = "";

$queryCondition = "";
if(!empty($_POST["search"])) {
    $advance_search_submit = $_POST["advance_search_submit"];
    foreach($_POST["search"] as $k=>$v){
        if(!empty($v)) {

            $queryCases = array("with_any_one_of","with_the_exact_of","without","starts_with");
            if(in_array($k,$queryCases)) {
                if(!empty($queryCondition)) {
                    $queryCondition .= " AND ";
                } else {
                    $queryCondition .= " WHERE ";
                }
            }
            switch($k) {
                case "with_any_one_of":
                    $with_any_one_of = $v;
                    $wordsAry = explode(" ", $v);
                    $wordsCount = count($wordsAry);
                    for($i=0;$i<$wordsCount;$i++) {
                        if(!empty($_POST["search"]["search_in"])) {
                            $queryCondition .= $_POST["search"]["search_in"] . " LIKE '%" . $wordsAry[$i] . "%'";
                        } else {
                            $queryCondition .= "question LIKE '" . $wordsAry[$i] . "%' OR answer LIKE '" . $wordsAry[$i] . "%'";
                        }
                        if($i!=$wordsCount-1) {
                            $queryCondition .= " OR ";
                        }
                    }
                    break;
                case "with_the_exact_of":
                    $with_the_exact_of = $v;
                    if(!empty($_POST["search"]["search_in"])) {
                        $queryCondition .= $_POST["search"]["search_in"] . " LIKE '%" . $v . "%'";
                    } else {
                        $queryCondition .= "question LIKE '%" . $v . "%' OR answer LIKE '%" . $v . "%'";
                    }
                    break;
                case "without":
                    $without = $v;
                    if(!empty($_POST["search"]["search_in"])) {
                        $queryCondition .= $_POST["search"]["search_in"] . " NOT LIKE '%" . $v . "%'";
                    } else {
                        $queryCondition .= "question NOT LIKE '%" . $v . "%' AND answer NOT LIKE '%" . $v . "%'";
                    }
                    break;
                case "starts_with":
                    $starts_with = $v;
                    if(!empty($_POST["search"]["search_in"])) {
                        $queryCondition .= $_POST["search"]["search_in"] . " LIKE '" . $v . "%'";
                    } else {
                        $queryCondition .= "question LIKE '" . $v . "%' OR answer LIKE '" . $v . "%'";
                    }
                    break;
                case "search_in":
                    $search_in = $_POST["search"]["search_in"];
                    break;
            }
        }
    }
}
$orderby = " ORDER BY quiz_id desc"; 
$sql = "SELECT * FROM questions " . $queryCondition;
$result = $db->exec($sql);
if($result)
{
    echo "Good";
    while ($row = $result->fetchArray(SQLITE3_ASSOC))
    {
        $question = $row['question'];
        echo $question;
    }
}
else 
{
    echo "No results found";
}
?>
<form name="frmSearch" method="post" action="searchtext.php">
<input type="hidden" id="advance_search_submit" name="advance_search_submit" value="<?php echo $advance_search_submit; ?>">
<div class="search-box">
    <label class="search-label">With Any One of the Words:</label>
    <div>
        <input type="text" name="search[with_any_one_of]" class="demoInputBox" value="<?php echo $with_any_one_of; ?>"  />
        <span id="advance_search_link" onClick="showHideAdvanceSearch()">Advance Search</span>
    </div>              
    <div id="advanced-search-box" <?php if(empty($advance_search_submit)) { ?>style="display:none;"<?php } ?>>
        <label class="search-label">With the Exact String:</label>
        <div>
            <input type="text" name="search[with_the_exact_of]" id="with_the_exact_of" class="demoInputBox" value="<?php echo $with_the_exact_of; ?>"   />
        </div>
        <label class="search-label">Without:</label>
        <div>
            <input type="text" name="search[without]" id="without" class="demoInputBox" value="<?php echo $without; ?>" />
        </div>
        <label class="search-label">Starts With:</label>
        <div>
            <input type="text" name="search[starts_with]" id="starts_with" class="demoInputBox" value="<?php echo $starts_with; ?>" />
        </div>
        <label class="search-label">Search Keywords in:</label>
        <div>
            <select name="search[search_in]" id="search_in" class="demoInputBox">
                <option value="">Select Column</option>
                <option value="title" <?php if($search_in=="title") { echo "selected"; } ?>>Title</option>
                <option value="description" <?php if($search_in=="description") { echo "selected"; } ?>>Description</option>
            </select>
        </div>
    </div>

    <div>
        <input type="submit" name="go" class="btnSearch" value="Search">
    </div>
</div>
</form>

请任何帮助将不胜感激。提前谢谢。

Answer 1

有几种不同的方法可以做到这一点。您可以尝试使用下面提到的方法 -

$query->fetch(SQLITE_ASSOC)，这是面向对象的方法。现在您正在遵循这种方法，使用这种方法更改您的代码，希望它能够正常工作。
另一种方法可以是像这段代码while ($entry = sqlite_fetch_array($query, SQLITE_ASSOC)) {

使用SQLite3数据库进行PHP搜索查询

1 个答案: