PHP变量不是mysqli对象

问题描述：

I have the next code (compilated in NetBeans)

$query = 'INSERT INTO grupos(nombre, descripcion)'
            . ' VALUES ("' . utf8_decode($_POST['name']) . '", "' . utf8_decode($_POST['desc']) . '")';
    $result = $con->query($query) or exit("Error: " . $con->errno . ": " . $con->error);
    if ($result->affected_rows > 0) {

Why I got this: "Trying to get property of non-object " if when I made a echo $result; display a '1'?

我有下一个代码（在NetBeans中编译） p>

   $ query ='INSERT INTO grupos（nombre，descripcion）'
。  'VALUES（“''。utf8_decode（$ _ POST ['name']）。'”，“'。utf8_decode（$ _ POST ['desc']）。'”）'; 
 $ result = $ con＆gt; query  （$ query）或退出（“错误：”。$ con＆gt; errno。“：”。$ con＆gt;错误）; 
 if（$ result-＆gt; affected_rows＆gt; 0）{
   code>  pre> 
 
 为什么我得到这个：“当我发出echo $结果时，”试图获得非对象的属性“
if; 显示'1'？   p> 
  div>

答

From the documentation of mysqli::query():

For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

Since your query is INSERT, not SELECT, it doesn't return a mysqli_result, it just returns TRUE. You should use $con->affected_rows instead of $result->affected_rows.

PHP变量不是mysqli对象

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