Question

我正在尝试使用php更新mySQL数据库，但我遇到了一些错误：

解析错误：语法错误，意外'}'，期待第29行C：\ xampp \ htdocs \ SLR \更新S110 PC01.php中的文件结束

以下代码：

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "slr";

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

$soft_id = $_POST['soft_id'];
$soft_name = $_POST['soft_name'];
$installed_date = $_POST['installed_date'];
$expiry_date = $_POST['expiry_date'];
$product_key = $_POST['product_key'];

$sql = "UPDATE s110_pc01 ". "SET soft_name = $soft_name ". 
              "WHERE soft_id = $soft_id" ;
$retval = mysql_query( $sql, $link );

if(! $retval ) {
               die('Could not update data: ' . mysql_error());
            }
            echo "Updated data successfully\n";
mysql_close($link);
 }else {
            ?>
<form method = "post" action = "<?php $_PHP_SELF ?>">
                  <table width = "400" border =" 0" cellspacing = "1" 
                     cellpadding = "2">

                     <tr>
                        <td width = "100">Software ID</td>
                        <td><input name = "soft_id" type = "text" 
                           id = "soft_id"></td>
                     </tr>

       <tr>
                        <td width = "100">Software Name</td>
                        <td><input name = "soft_name" type = "text" 
                           id = "soft_name"></td>
                     </tr>

Answer 1

尝试下面

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "slr";

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

$soft_id = $_POST['soft_id'];
$soft_name = $_POST['soft_name'];
$installed_date = $_POST['installed_date'];
$expiry_date = $_POST['expiry_date'];
$product_key = $_POST['product_key'];

$sql = "UPDATE s110_pc01 ". "SET soft_name = $soft_name ". 
              "WHERE soft_id = $soft_id" ;
$retval = mysql_query( $sql, $link );


if($retval ) {
    echo "Updated data successfully\n";       
} else {
die('Could not update data: ' . mysql_error());          
mysql_close($link);
}
?>
<form method = "post" action = "<?php $_PHP_SELF ?>">
                  <table width = "400" border =" 0" cellspacing = "1" 
                     cellpadding = "2">

                     <tr>
                        <td width = "100">Software ID</td>
                        <td><input name = "soft_id" type = "text" 
                           id = "soft_id"></td>
                     </tr>

       <tr>
                        <td width = "100">Software Name</td>
                        <td><input name = "soft_name" type = "text" 
                           id = "soft_name"></td>
                     </tr>

使用php更新mysql数据库时出错

1 个答案: