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php - 减法时如何用php更新mysql

转载 作者:行者123 更新时间:2023-11-29 00:45:54 25 4
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我想用 PHP 和 mySQL 更新一个表。我需要添加或减去我从 mysql 数据库中已有的值中选择的任何数字。当我添加到“player_qty”时它工作正常但当我减去/删除时它不工作。它给了我我想要的消息,很多已经被删除,但是当我检查数据库时,值是相同的——不是删除。

$register_query = "SELECT * FROM players WHERE player_id='$add_player_id' AND event_name = '$event_name'";
$register_check = mysql_query($register_query, $db);
$register_found_users = mysql_num_rows($register_check);

if($_POST['button'] == "Add") {
if($register_found_users>0){
mysql_query("UPDATE players
SET
player_qty = player_qty + $add_player_qty,
player_pickedup = '1',
event_user = '$event_user'
WHERE player_id = '$add_player_id'");
$msg = $add_player_qty.' more ticket(s) added to '.$add_player_id.'.';
} else {
$insertSQL1 = "
INSERT INTO players
(player_id, player_qty, player_adt, player_level, player_pickedup, event_name, event_type, event_end, event_user)
VALUES
('$add_player_id', '$add_player_qty', NULL, NULL, '1', '$event_name', '$event_type', '$event_end', '$event_user')";
mysql_select_db($dbname, $db);
$Result1 = mysql_query($insertSQL1, $db) or die(mysql_error());
$msg = $add_player_id.' added to DB w/ '.$add_player_qty.' ticket(s)';
}
}elseif($_POST['button'] == "Delete") {
if($register_found_users>0){
mysql_query("UPDATE players
SET
player_qty = player_qty - $add_player_qty,
player_pickedup = '1',
event_user = $event_user
WHERE player_id = '$add_player_id'");
$msg = $add_player_qty.' ticket(s) deleted from '.$add_player_id.'.';
} else {
$msg = $add_player_id.' not in database. Please add manually.';
}
} else {
//
}
<!-- FORM TO ADD PLAYER TO DATABASE IF DOESN'T EXIST -->
<form action="" method="post" enctype="multipart/form-data" name="form_add_player" id="form_add_player" style="display:<?php echo $css;?>;">

<label for="add_player_id">PLAYER #: </label>
<input type="text" name="add_player_id" id="add_player_id" value=""/>

<label for="add_player_qty">QUANTITY: </label>
<select name="add_player_qty" id="add_player_qty">
<option value="1">1</option>
<option value="2" selected>2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
<option value="6">6</option>
<option value="7">7</option>
<option value="8">8</option>
<option value="9">9</option>
<option value="10">10</option>
</select>

<input name="event_end" type="hidden" id="event_end" value="<?php echo $row_get_type['event_end'];?>" />
<input type="submit" name="button" value="Add" class="button">
<input type="submit" name="button" value="Delete" class="button">

</form>

最佳答案

在您的减法查询中,您没有在您的输入值之一上设置 ' - 也许这会导致 mysql 错误,因此查询不会被执行。

mysql_query("UPDATE `players` 
SET
`player_qty` = `player_qty` - $add_player_qty,
`player_pickedup` = '1',
`event_user` = '$event_user'
WHERE `player_id` = '$add_player_id'");

关于php - 减法时如何用php更新mysql,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10536929/

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