Question

我正在使用一个简单的表单在我的数据库中的表上运行数据库查询。连接似乎不是问题。表单呈现没有任何问题。当我进入页面时，我收到一个错误，指出我正在尝试获取非对象的属性。这是被调用的特定行：

if ($result->num_rows > 0){
    echo "$result";
}

任何想法为什么？

<?php

$servername = "localhost";
$username = "xxx";
$password = "xxx";
$dbname = "oldga740_SeniorProject";


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


$result= '';
//collect 
if(isset($_POST['search'])) {
  $searchq = $_POST['search'];
  $searchq = preg_replace("#[^0-9a-z]#i","",$searchq);

  $sql = "SELECT * FROM Customer WHERE Client LIKE '%$searchq%'";
  $result = $conn->query($sql);
}

//Display results
if ($result->num_rows > 0) {
     echo "$result";

}

else {
     echo "0 results";
}


?>

<html>

<head>

</head>

<body>

<form action="Index.php" method="post">
  <input type="text" name="search" placeholder="Search...." />
  <input type="submit" value=">>" />
</form>



</body>

</html>

Answer 1

这是相当直接的。我已经注意到了清晰度：

// You make it a variable here, with the assumption
// $_POST['search'] will transform it into an object later
$result= '';
// If there is a search variable try to search database
if(isset($_POST['search'])) {
    $searchq = $_POST['search'];
    $searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
    $sql = "SELECT * FROM Customer WHERE Client LIKE '%$searchq%'";

    // All is good if the condition is met
    $result = $conn->query($sql);

    // This has to go here because you have turned $result into an object now
    if ($result->num_rows > 0) {
        // Likely you will draw an error here, this is probably an array
        // that you will need to iterate over using while()
        echo "$result";
    }
}

/*
// If you leave it here, if the search is not being done, you've
// assigned $result = '' so you are doing ->num_rows on empty
if ($result->num_rows > 0) {
     echo "$result";

}
*/

错误消息在尝试显示查询结果时尝试获取非对象的属性

1 个答案: