Question

<?php
$dbhost = 'localhost';
$dbuser = 'santhosh';
$dbpass = 'Santhu';
$database = 'car';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($database, $conn);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
$name="nair";
$lat="123";
$lon="456";
$sql = "UPDATE cart SET latitude='$lat',longitude='$lon' WHERE uname<="$name";
if (mysql_query($sql)) {
echo "New record updated successfully";
} else {
echo "Error: " ;
}
mysql_close($conn);
?>

我收到消息＆＃34;错误＆＃34;这意味着我无法更新我的表格。请告诉我我的代码中有什么错误...先谢谢你

Answer 1

请更改您的查询，如下所示。

"UPDATE cart SET latitude='$lat',longitude='$lon' WHERE uname = '".$name."'";

我希望这会对你有所帮助

Answer 2

我认为查询会像这样

  $sql = "UPDATE `cart` SET `latitude`='".$lat."',`longitude`='".$lon."' WHERE `uname`='".$name."'";

Answer 3

<?php
$dbhost = 'localhost';
$dbuser = 'santhosh';
$dbpass = 'Santhu';
$database = 'car';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($database, $conn);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
$name="nair";
$lat="123";
$lon="456";
$sql = "UPDATE cart SET latitude='$lat',longitude='$lon' WHERE uname='$name'";
if (mysql_query($sql)) {
echo "New record updated successfully";
} else {
echo "Error: " ;
}
mysql_close($conn);
?>

Answer 4

您的查询中有一个错误引用的错误。

"UPDATE cart SET latitude='$lat',longitude='$lon' WHERE uname<=$name";
                                                             ^^^^^^^^

更新了查询

"UPDATE cart SET latitude = '$lat',longitude = '$lon' WHERE uname = '".$name."'";
                                                                  ^^^^^^^^^^^^^^

Answer 5

$sql = "UPDATE cart SET latitude = '$lat',longitude='$lon' WHERE uname = '".$name."'";

这段代码解决了我的问题。谢谢大家的答案......

无法使用php更新mysql表

5 个答案: