Question

我正在制作一个用于博客出版物的CRUD系统，但它有点奇怪，另一个开发人员（有更多经验）看着我的编码，对他来说也没关系，但这个错误（注意：试图获得财产第32行的C：\ xampp \ htdocs \ genial \ painel \ inc \ database.php中的非对象）仍然出现。

我的数据库代码：

Uncaught ReferenceError: IframeResizer is not defined

很抱歉，如果它没有被闲置。

我在

的“FROM”后面的代码上放了一个空格

<?php
 mysqli_report(MYSQLI_REPORT_STRICT);
function open_database() {
try {
    $conn = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
    return $conn;
} catch (Exception $e) {
    echo $e->getMessage();
    return null;
 }
}
function close_database($conn) {
try {
    mysqli_close($conn);
} catch (Exception $e) {
    echo $e->getMessage();
    }
}

function find( $table = null, $id = null ) {

    $database = open_database();
    $found = null;
    try {
        if ($id) {
            $sql = "SELECT * FROM" . $table . " WHERE id = " . $id;
            $result = $database->query($sql);

            if ($result->num_rows > 0) {
                $found = $result->fetch_assoc();
                }

            } 
           else {

              $sql = "SELECT * FROM " . $table;
              $result = $database->query($sql);

             if ($result->num_rows > 0) {
                $found = $result->fetch_all(MYSQLI_ASSOC);
             }
    }
} catch (Exception $e) {
    $_SESSION['message'] = $e->GetMessage();
    $_SESSION['type'] = 'danger';
   }

close_database($database);
return $found;
}



 function find_all( $table ) {
    return find($table);
 }

 function save($table = null, $data = null) {
     $database = open_database();
     $columns = null;
     $values = null;
     //print_r($data);
     foreach ($data as $key => $value) {
        $columns .= trim($key, "'") . ",";
        $values .= "'$value',";
     }

     $columns = rtrim($columns, ',');
     $values = rtrim($values, ',');

     $sql = "INSERT INTO " . $table . "($columns)" . " VALUES " ($values);";
     try {
        $database->query($sql);
        $_SESSION['message'] = 'Registro cadastrado com sucesso.';
        $_SESSION['type'] = 'success';

    } catch (Exception $e) {

        $_SESSION['message'] = 'Nao foi possivel realizar a operacao.';
        $_SESSION['type'] = 'danger';
        }
      close_database($database);
    }

    function update($table = null, $id = 0, $data = null) {
        $database = open_database();
        $items = null;
        foreach ($data as $key => $value) {
            $items .= trim($key, "'") . "='$value',";
        }

        $items = rtrim($items, ',');
        $sql  = "UPDATE " . $table;
        $sql .= " SET $items";
        $sql .= " WHERE id=" . $id . ";";
        try {
            $database->query($sql);
            $_SESSION['message'] = 'Registro atualizado com sucesso.';
            $_SESSION['type'] = 'success';
        }
        catch (Exception $e) {
            $_SESSION['message'] = 'Nao foi possivel realizar a operacao.';
            $_SESSION['type'] = 'danger';
        }
        close_database($database);
    }

错误保持不变，但在第36行知道：

$sql = "SELECT * FROM" . $table . " WHERE id = " . $id;

将第36行的代码转为：

if ($result->num_rows > 0) {
    $found = $result->fetch_all(MYSQLI_ASSOC);
}

错误消失，其他问题与此问题无关。

Answer 1

行$sql = "SELECT * FROM" . $table . " WHERE id = " . $id;在FROM之后是否有空格。

Answer 2

如果此问题未被off-topic -> typo generated关闭，我最好提供可接受的答案。

在FROM查询中的$table和SELECT之间添加空格。
检查查询中的非错误结果。

您的新代码可能如下所示：

$sql="SELECT * FROM `$table`";
if($id){
    // assuming id has been properly sanitized
    $sql.=" WHERE id=$id";  // concatenate with WHERE clause when appropriate
}
if($result=$database->query($sql)){  // only continue if result isn't false
    if($result->num_rows){  // only continue if one or more row
        $found=$result->fetch_all(MYSQLI_ASSOC);  // fetch all regardless of 1 or more
    }
}

试图获得非对象CRUD的属性

2 个答案: