可捕获的致命错误：类mysqli_result的对象无法在第73行转换为字符串

Question

我正在制作一个用于更新数据库中数据的表单，但它给我发现了这个错误Catchable fatal error: Object of class mysqli_result could not be converted to string on line 73

我尝试输出所有数据，但都在打印，但没有在数据库中更新。

请提出一些建议，如何解决？

$id = $_GET['id'];
$t_id = $_GET['table_id'] - 2;

if(isset($_POST['submit'])){

  $name = $_POST['name'];
  $phone = $_POST['phone'];
  $email = $_POST['email'];
  $gender = intval($_POST['gender']);
  $treatment = intval($_POST['treatment']);
  $source = intval($_POST['source']);
  $status_ = intval($_POST['status']);
  $remark = $_POST['remark'];

//below code is line 73

$leadUpdateSql = "UPDATE lead SET
                     name = $name
                    ,phone = $phone
                    ,email = $email
                    ,gender_id = $gender
                    ,treatment_id = $treatment
                    ,source_id = $source
                    ,status_id = $status
                    ,remark = $remark
                     WHERE lead.id = $id";

if ($conn->query($leadUpdateSql) === TRUE) {
    echo "Record updated successfully";
    header("location:edit.php?id=$id&table_id=$t_id&updated=successfully");
   } 

}

Answer 1

使用email = '.$email.'制作SQL。您最好打印SQL字符串并在MySQL CMD或PHPMyAdmin中执行以查找。

Answer 2

我设法解决了这个问题。我为字符串值添加了引号。

$id = $_GET['id'];
$t_id = $_GET['table_id'] - 2;

if(isset($_POST['submit'])){

    $name = $_POST['name'];
    $phone = $_POST['phone'];
    $email = $_POST['email'];
    $gender = intval($_POST['gender']);
    $treatment = intval($_POST['treatment']);
    $source = intval($_POST['source']);
    $status_ = intval($_POST['status']);
    $remark = $_POST['remark'];



$leadUpdateSql = "UPDATE lead SET
         name='$name' // here was the issue, now added quotes
        ,phone='$phone' // here was the issue
        ,email='$email' // here was the issue
        ,gender_id=$gender
        ,treatment_id=$treatment
        ,source_id=$source
        ,status_id=$status_
        ,remark='$remark'   // here was the issue
         WHERE lead.id=$id";

if ($conn->query($leadUpdateSql) === TRUE) {

    header("location:index.php?updated=successfully&table_id=$table_id#$t_id");
   }

else{
    header("location:edit.php?id=$id&table_id=$t_id&error=unsuccess");
}
}