无法清除错误 mysql_fetch_array() 期望参数 1 是资源，给定对象答案

【问题标题】：Can't clear error mysql_fetch_array() expects parameter 1 to be resource, object given无法清除错误 mysql_fetch_array() 期望参数 1 是资源，给定对象
【发布时间】：2015-12-01 04:15:41
【问题描述】：

我正在尝试构建一个 SELECT 查询，该查询允许我使用返回结果的表单来处理对选定行的 UPDATE。我收到的警告出现在我的代码的第 36 行。我以为我正确设置了代码来构建表格，然后填充它，但我找不到我的错误。我看到了这个问题的几个版本，但我没有看到一个谈到给出的对象。任何帮助都将受到欢迎。

<html>
<head>
</head>
<body>
<?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);
}
$sql = "SELECT * FROM Projects";
$result = $conn->query($sql);
echo "<table border=1>
<tr>
<th>Project</th>
<th>Client</th>
<th>Last Name</th>
<th>Date Received</th>
<th>Final Review Date</th>
<th>Date Delivered</th>
<th>Date Accepted>
</tr>";
while($record = mysql_fetch_array($result))
{
if ($result->num_rows > 0){
echo "<form action=mynewform.php method=post>";
echo "<tr>"; 
echo "<td>" . "<input type=text name=project value=" . $record['Project'] . " </td>";
echo "<td>" . "<input type=text name=client value=" . $record['Client'] . " </td>";
echo "<td>" . "<input type=text name=lastname value=" . $record['LastName'] . " </td>";
echo "<td>" . "<input type=text name=datereceived value=" . $record['DateReceived'] . " </td>";
echo "<td>" . "<input type=text name=finalreview date value=" . $record['FinalReviewDate'] . " </td>";
echo "<td>" . "<input type=text name=datedelivered value=" . $record['DateDelivered'] . " </td>";
echo "<td>" . "<input type=text name=dateaqccepted value=" . $record['DateAccepted'] . " </td>";
echo "<td>" . "<input type=hidden name=hidden value=" . $record['Project'] . " </td>";
echo "<td>" . "<input type=submit name=update value=update" . " </td>";
echo "<td>" . "<input type=submit name=delete value=delete" . " </td>";
echo "</tr>";
echo "</form>";
}
}
echo "</table>";
if (isset($_POST[update])){
$UpdateQuery = "UPDATE Projects SET Project='$_POST[project]', Client='$_POST[client]', LastName='$_POST[lastname]', DateReceived='$_POST[datereceived]', FinalReviewDate='$_POST[finalreviewdate]', DateDelivered='$_POST[datedelivered]', DateAccepted='$_POST[dateaccepted]' WHERE Project='$_POST[hidden]";
mysql_query($UpdateQuery, $con);
}; 
?>
<?php
    $connection->close();
?>
</body>
</html>

编辑代码

<html>

<head>

</head>

<body>

<?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);
}

$sql = "SELECT * FROM Projects";
$result = $conn->query($sql);

echo "<table border=1>
<tr>
<th>Project</th>
<th>Client</th>
<th>Last Name</th>
<th>Date Received</th>
<th>Final Review Date</th>
<th>Date Delivered</th>
<th>Date Accepted</th>
</tr>";

while($record = mysqli_fetch_array($result))
{
if ($result->num_rows > 0){

echo "<form action='mynewform'.php method='post'>";
echo "<tr>"; 
echo "<td>" . "<input type='text' name='project' value='" . $record['Project'] . " </td>";
echo "<td>" . "<input type='text' name='client' value='" . $record['Client'] . " </td>";
echo "<td>" . "<input type='text' name='lastname' value='" . $record['LastName'] . " </td>";
echo "<td>" . "<input type='text' name='datereceived' value='" . $record['DateReceived'] . " </td>";
echo "<td>" . "<input type='text' name='finalreviewdate' value='" . $record['FinalReviewDate'] . " </td>";
echo "<td>" . "<input type='text' name='datedelivered' value='" . $record['DateDelivered'] . " </td>";
echo "<td>" . "<input type='text' name='dateaccepted' value='" . $record['DateAccepted'] . " </td>";
echo "<td>" . "<input type='hidden' name='hidden' value='" . $record['Project'] . " </td>";
echo "<td>" . "<input type='submit' name='update' value=update'" . " </td>";
echo "<td>" . "<input type='submit' name='delete' value=delete'" . " </td>";
echo "</tr>";
echo "</form>";
}
}
echo "</table>";

if (isset($_POST['update'])){
$UpdateQuery = "UPDATE Projects SET Project='$_POST[project]', Client='$_POST[client]', LastName='$_POST[lastname]', DateReceived='$_POST[datereceived]', FinalReviewDate='$_POST[finalreviewdate]', DateDelivered='$_POST[datedelivered]', DateAccepted='$_POST[dateaccepted]' WHERE Project='$_POST[hidden]";
mysqli_query($conn, $sql);
}; 

?>

<?php
    $conn->close();
?>

</body>

</html>

第二次编辑

<html>

<head>

</head>

<body>

<?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);
}

$sql = "SELECT * FROM Projects";
$result = $conn->query($sql);

echo "<table border=1>
<tr>
<th>Project</th>
<th>Client</th>
<th>Last Name</th>
<th>Date Received</th>
<th>Final Review Date</th>
<th>Date Delivered</th>
<th>Date Accepted</th>
</tr>";

while($record = mysqli_fetch_array($result))
{
if ($result->num_rows > 0){

echo "<form action='mynewform.php' method='post'>";
echo "<tr>"; 
echo "<td>" . "<input type='text' name='project' value='" . $record['Project'] . "' /></td>";
echo "<td>" . "<input type='text' name='client' value='" . $record['Client'] . "'/></td>";
echo "<td>" . "<input type='text' name='lastname' value='" . $record['LastName'] . "' /></td>";
echo "<td>" . "<input type='text' name='datereceived' value='" . $record['DateReceived'] . "' /></td>";
echo "<td>" . "<input type='text' name='finalreviewdate' value='" . $record['FinalReviewDate'] . "' /></td>";
echo "<td>" . "<input type='text' name='datedelivered' value='" . $record['DateDelivered'] . "' /></td>";
echo "<td>" . "<input type='text' name='dateaccepted' value='" . $record['DateAccepted'] . "' /></td>";
echo "<td>" . "<input type='hidden' name='hidden' value='" . $record['Project'] . "' /></td>";
echo "<td>" . "<input type='submit' name='update' value=update'" . "' /></td>";
echo "<td>" . "<input type='submit' name='delete' value=delete'" . "' /></td>";
echo "</tr>";
echo "</form>";
}
}
echo "</table>";

if (isset($_POST['update'])){
$UpdateQuery = "UPDATE Projects SET Project='$_POST[project]', Client='$_POST[client]', LastName='$_POST[lastname]', DateReceived='$_POST[datereceived]', FinalReviewDate='$_POST[finalreviewdate]', DateDelivered='$_POST[datedelivered]', DateAccepted='$_POST[dateaccepted]' WHERE Project='$_POST[hidden]";
mysqli_query($conn, $sql);
}; 

?>

<?php
    $conn->close();
?>

</body>

</html>

【问题讨论】：

您有一些错误。一种是您将mysql_ 驱动程序与mysqli 混合使用。另一个是 $connection 未定义。您也对 SQL 注入持开放态度。 a) mysql_fetch_array b) mysql_query
此外，您的 HTML 元素无效，您缺少结束标签，您的属性应该被引用。（不是当前的问题，但一旦你得到 SQL 处理就会出现）。
是的，这是我提到的第一个错误； you are mixing mysql_ driver with mysqli 所以$result 是一个mysqli 结果对象； mysql_ 函数无法使用它。看看php.net/manual/en/mysqli-result.fetch-array.php。
$result 是 mysqli 查询，所以不能在 while($record = mysql_fetch_array($result)) 中使用 mysql_fetch_array($result)。试试while($record = $result->fetch_array())
您需要引用键，否则 php 将假定它们是常量。所以把if (isset($_POST[update])){改成if (isset($_POST['update'])){（即'update'）

标签： php mysqli

【解决方案1】：

除了之前在 cmets 中的注释。我没有在您的 HTML 元素上看到任何其他答案/cmets，因此我将在这些元素上添加答案。

您将使用此 PHP 生成的 HTML 将无效。属性需要引用，输入元素需要关闭。

例如你当前的代码

echo "<form action=mynewform.php method=post>";
echo "<tr>"; 
echo "<td>" . "<input type=text name=project value=" . $record['Project'] . " </td>";

会输出：

<form action=mynewform.php method=post>
<tr>
<td><input type=text name=project value=what ever projects value is</td>

这是无效的 element 这里名为 project 永远不会关闭，它的值将只是 what。（或在某些浏览器中；由于标记无效，应该会出现意外结果）

你的 PHP 应该是：

echo "<form action='mynewform.php' method='post'>";
echo "<tr>"; 
echo "<td>" . "<input type='text' name='project' value='" . $record['Project'] . "' /> </td>";

这应该给你

<form action='mynewform.php' method='post'>
<tr>
<td><input type='text' name='project' value='what ever projects value is' /></td>

请注意第二个示例与第一个示例中的语法突出显示有何不同。

另外请查看http://php.net/manual/en/mysqli.quickstart.prepared-statements.php，或者安全性稍差但比您目前拥有的方法更好的方法http://php.net/manual/en/mysqli.real-escape-string.php。

有关整个主题的详细信息，请参阅：
a) How can I prevent SQL injection in PHP?
b) http://php.net/manual/en/security.database.sql-injection.php
c) https://www.owasp.org/index.php/SQL_Injection_Prevention_Cheat_Sheet

更新：

这是我要采用的结构（除了 SQL 注入孔）：

<?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: " . $conn->connect_error);
}
if (isset($_POST['update'])){
    $UpdateQuery = "UPDATE Projects SET Project='$_POST[project]', Client='$_POST[client]', LastName='$_POST[lastname]', DateReceived='$_POST[datereceived]', FinalReviewDate='$_POST[finalreviewdate]', DateDelivered='$_POST[datedelivered]', DateAccepted='$_POST[dateaccepted]' WHERE Project='$_POST[hidden]";
    mysqli_query($conn, $sql);
}
$sql = "SELECT * FROM Projects";
$result = $conn->query($sql);
?>
<html>
    <head>
    </head>
    <body>
        <table border='1'>
            <tr>
                <th>Project</th>
                <th>Client</th>
                <th>Last Name</th>
                <th>Date Received</th>
                <th>Final Review Date</th>
                <th>Date Delivered</th>
                <th>Date Accepted</th>
            </tr>
<?php
if ($result->num_rows > 0){
     while($record = mysqli_fetch_array($result)) {?>
            <form action='mynewform.php' method='post'>
                <tr>
                    <td><input type='text' name='project' value='<?php echo $record['Project'];?>' /></td>
                    <td><input type='text' name='client' value='<?php echo $record['Client'];?>'/></td>
                    <td><input type='text' name='lastname' value='<?php echo $record['LastName'];?>' /></td>
                    <td><input type='text' name='datereceived' value='<?php echo $record['DateReceived'];?>' /></td>
                    <td><input type='text' name='finalreviewdate' value='<?php echo $record['FinalReviewDate'];?>' /></td>
                    <td><input type='text' name='datedelivered' value='<?php $record['DateDelivered'];?>' /></td>
                    <td><input type='text' name='dateaccepted' value='<?php echo $record['DateAccepted'];?>' /></td>
                    <td><input type='hidden' name='hidden' value='<?php echo $record['Project'];?>' /></td>
                    <td><input type='submit' name='update' value='update' /></td>
                    <td><input type='submit' name='delete' value='delete' /></td>
                </tr>
            </form>
<?php
    }
}
$conn->close();
?>
        </table>
    </body>
</html>

【讨论】：

克里斯，感谢您的额外意见。当我看到这条评论时，我开始研究为什么我的表格没有产生任何结果。我知道你要这样做了，但是当我进行更改时，表格没有正确呈现。
您能否更新您的问题以显示您所做的更改？
您错过了关闭输入元素。请参阅我的回答中的. "' /> </td>"。对比你的； . " </td>".
好的，表单又回来了，但我在更新和删除按钮中显示“，它仍然不会将任何内容传递给更新查询。感谢您尝试获取此信息的所有帮助解决了，但我不希望你花一整天的时间来帮助我解决这个问题。
请用您更新的代码更新问题。您的更新查询无效，您缺少WHERE Project='$_POST[hidden] 上的结束'。这仍然对 SQL 注入开放。我看不到您在哪里使用删除。您应该对这些查询以及页面本身使用错误报告。

【解决方案2】：

将mysql_fetch_array($result) 更改为$result->fetch_array()

     if ($result->num_rows > 0){



   while($record = $result->fetch_array())
    {

    echo "<form action=mynewform.php method=post>";
    echo "<tr>"; 
    echo "<td>" . "<input type=text name=project value=" . $record['Project'] . " </td>";
    echo "<td>" . "<input type=text name=client value=" . $record['Client'] . " </td>";
    echo "<td>" . "<input type=text name=lastname value=" . $record['LastName'] . " </td>";
    echo "<td>" . "<input type=text name=datereceived value=" . $record['DateReceived'] . " </td>";
    echo "<td>" . "<input type=text name=finalreview date value=" . $record['FinalReviewDate'] . " </td>";
    echo "<td>" . "<input type=text name=datedelivered value=" . $record['DateDelivered'] . " </td>";
    echo "<td>" . "<input type=text name=dateaqccepted value=" . $record['DateAccepted'] . " </td>";
    echo "<td>" . "<input type=hidden name=hidden value=" . $record['Project'] . " </td>";
    echo "<td>" . "<input type=submit name=update value=update" . " </td>";
    echo "<td>" . "<input type=submit name=delete value=delete" . " </td>";
    echo "</tr>";
    echo "</form>";
    }
    }

你不能合并mysql和mysqli

【讨论】：

将if ($result->num_rows > 0) 移到while($record = $result->fetch_array()) 之外会更有意义。

【解决方案3】：

你用mysqli_连接MYSQL，想得到结果的时候还需要mysqli_。

这是来自官方的示例：

`$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* check connection */
if ($mysqli->connect_errno){
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}
$sql = "SELECT a.uid, a.role AS roleid, b.role,FROM tbl_usr a INNER JOIN tbl_memrole b ON a.role = b.id  ";
    if ($result = $mysqli->query($sql)) { 
        while($obj = $result->fetch_object()){ 
            $line.=$obj->uid; 
            $line.=$obj->role; 
            $line.=$obj->roleid; 
        }
    }
    $result->close(); 
    unset($obj); 
    unset($sql); 
    unset($query);`

【讨论】：