Question

我的mySQLi查询失败，但错误是空白的。任何人都可以告诉我问题是什么，或者我如何输出错误信息？

继承我的PHP代码：

// Create connection
$conn = new mysqli($hostname, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

if(isset($_POST['in_wordarray']))       
{       
    $words = $_POST['in_wordarray'];

    $sql = "SELECT *, (";

    $i = 0;
    foreach ($words as $value) {

        if($i == 0) {
            $sql .= "(`words` LIKE '%$value%')";
        } else {
            $sql .= " + (`words` LIKE '%$value%')";
        }

        $i++;
    }

    $sql .= ") AS `numMatches` FROM `mytable` HAVING `numMatches` >= 3 ORDER BY `numMatches` DESC";
    //echo $sql;
    $result = $conn->query($sql);

    if ($result === TRUE) {
        $text_result_array = array();
        if ($result->num_rows > 0) {
          while ($row = $result->fetch_assoc()) {
              $text_result_array[] = $row;
          }
        // Encode the response
        echo json_encode($text_result_array);
        }
    } else {
        echo $conn->error;
    }
} else {
    echo "Bad Input";

}

生成的SQL查询如下所示：

SELECT *, ((`words` LIKE '%word1%') + (`words` LIKE '%word2%') + (`words` LIKE '%word3%')) AS `numMatches` FROM `words`HAVING `numMatches` >= 3 ORDER BY `numMatches` DESC

Answer 1

抱歉，我没有单独使用php的经验，但你确定这会返回一个布尔值吗？

$result = $conn->query($sql);

如果没有，你可以删除if语句吗？

我把它写成答案，因为遗憾的是我还没有发表评论。

PHP / MYSQL查询失败但没有错误

1 个答案: