显示查询结果的PDO错误:PDOStatement类的对象无法转换为字符串

时间:2013-01-10 19:52:27

标签: php ajax

所以我第一次尝试AJAX而且我很激动!我有一个名为ajaxtest1.html的文件,它只有一个简单的文本框,其中输入里程,然后传递到另一个接收它的文件,将其插入查询并发回结果。这是ajaxtest1 ......

<html>
<body>
<script language="javascript" type="text/javascript">// <![CDATA[
function ajaxFunction(){
var ajaxRequest;  

try{
    // Opera 8.0+, Firefox, Safari
    ajaxRequest = new XMLHttpRequest();
} catch (e){
    // Internet Explorer Browsers
    try{
        ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
    } catch (e) {
        try{
            ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
        } catch (e){
            // Something went wrong
            alert("Your browser broke!");
            return false;
        }
    }
}
// receive data sent from the server
ajaxRequest.onreadystatechange = function(){
    if(ajaxRequest.readyState == 4){
        document.myForm.time.value = ajaxRequest.responseText;
    }
}
var miles = document.getElementById('miles').value;
var queryString = "?miles=" + miles;
ajaxRequest.open("GET", "results.php" + queryString, true);
ajaxRequest.send(null); 
}
// ]]></script>
<form name="myForm">Miles: <input id="miles" type="text" /> <br /> <br /> <input onclick="ajaxFunction()" value="Query MySQL" type="button" /></form>
</body>
</html>

然后,在results.php中。我有

<?php
$hostname = 'host';


$username = 'user';

$password = 'password';


$miles = $_GET['miles'];

try {
    $dbh = new PDO("mysql:host=$hostname;dbname=ratetable", $username, $password);
    echo 'Connected to database<br />';
    $stmt = $dbh->prepare('SELECT * FROM rates WHERE mileage <= :miles ORDER BY mileage DESC LIMIT 1');
    $stmt->bindParam(':miles', $miles, PDO::PARAM_INT);
    if ($stmt->execute()) {

    // get the rowcount
    $numrows = $stmt->rowCount();
    if ($numrows > 0) {
    // match
    // Fetch rows
      $rowset = $stmt->fetchAll();
    }
    else {
    // no rows
    }
   }

$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Mileage</th>";
$display_string .= "<th>Rate Per Mile</th>";
$display_string .= "<th>Inbound skid rate</th>";
$display_string .= "<th>Inbound truckload</th>";
$display_string .= "</tr>";

while($row_object = array($miles)){
$display_string .= "<tr>";
$display_string .= "<td>$row[mileage]</td>";
$display_string .= "<td>$row[ratepermile]</td>";
$display_string .= "<td>$row[skidinbound]</td>";
$display_string .= "<td>$row[truckinbound]</td>";
$display_string .= "</tr>";

   }
    echo "Query: " . $stmt . "<br />";
    $display_string .= "</table>";
    echo $display_string;


/*** close the database connection ***/
    $dbh = null;
}
catch(PDOException $e)
    {
    echo $e->getMessage();
    }
?>

[增订]

我很擅长使用PDO(好吧,PHP一般而且我正在接受它。)

我收到以下错误:

Catchable fatal error: Object of class PDOStatement could not be converted to string

这发生在以下行:

echo "Query: " . $stmt . "<br />";

它正好连接数据库。有人可以告诉我可能有什么问题吗?在试图形成PDO语句时,我有点想法,因此它会处理我的查询,并且它在最后一点上窒息。

我已经更新了PHP代码,以向您展示我尝试过的内容。

感谢您一看。

2 个答案:

答案 0 :(得分:0)

在results.php文件中尝试此操作:

$stmt = $dbh->prepare('SELECT * FROM rates WHERE mileage <= :miles ORDER BY mileage DESC LIMIT 1');
$stmt->bindParam(':miles', $miles, PDO::PARAM_INT);

你可以从这一行删除演员阵容:

$miles = (int) $_GET['miles'];

答案 1 :(得分:0)

“可捕获的致命错误:PDOStatement类的对象无法转换为字符串。”完全意味着它所说的。 $ stmt变量是一个PDO对象,你试图像它是一个字符串一样回应它。你不能回显一个对象,就像你不能只回显一个数组一样。

替换此行

$stmt = $dbh->prepare('SELECT * FROM rates WHERE mileage <= ? ORDER BY mileage DESC LIMIT 1');

有了这个:

$query_text = 'SELECT * FROM rates WHERE mileage <= ? ORDER BY mileage DESC LIMIT 1';
$stmt = $dbh->prepare($query_text);

然后替换此行

echo "Query: " . $stmt . "<br />";

有了这个

echo "Query: " . $query_text . "<br />";