警告:mysqli_real_escape_string()期望参数1为mysqli,字符串给定 [英] Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, string given
问题描述
我的网站有一个安全区域,可以添加/更新/删除数据库条目。我正在尝试将脚本从mysql更新到mysqli。一切都有效,除了更新部分。当我填写新信息并点击更新时,它会给我这个错误:
警告:mysqli_real_escape_string()期望参数1到
字符串在
/home3/tarb89/public_html/aususrpg.net/charbase/updated.php在线19中给出
我不确定这里有什么问题;我是一个完整的新手,所以我很抱歉。
这是我的update.php代码:
<? php
//创建连接
$ con = mysqli_connect(xxx,xxx,xxx,xxx);
//检查连接
if(mysqli_connect_errno())
{
echo无法连接到MySQL:。 mysqli_connect_error();
}
$ id = $ _GET ['id'];
$ result = mysqli_query($ con,SELECT * FROM characters WHERE id ='$ id');
$ my_array = array($ c_z);
extract($ my_array);
?>
< form id =FormNameaction =../ charbase / updated.phpmethod =postname =FormName>
< table width =448border =0cellspacing =2cellpadding =0>
< tr>
< td width =150align =right>< label for =name>名称:< / label>< / td>
< td>< input name =namemaxlength =255type =textvalue =<?php echo stripslashes($ name)?>>< / td>
< / tr>
< tr>
< td colspan =2align =center>
< input name =type =submitvalue =Update>
< input name =idtype =hiddenvalue =<?php echo $ id?>>
< / td>
< / tr>
< / table>
< / form>
我对update.php页面的另一个问题是当它显示表时,输入应该是显示已在数据库中列出的信息;目前,它显示名称字段,但输入部分为空。它应该显示数据库中已有的名称数据(然后我可以将其替换为我想要更新的任何内容。)
这是updated.php代码:
<?php
//创建连接
$ con = mysqli_connect(xxx,xxx , XXX, XXX);
//检查连接
if(mysqli_connect_errno())
{
echo无法连接到MySQL:。 mysqli_connect_error();
}
$ id = $ _POST ['id'];
$ name = mysqli_real_escape_string(trim($ _ POST [name]),$ con);
$ rsUpdate = mysqli_query($ con,UPDATE characters SET name ='$ name'
WHERE id ='$ id');
if($ rsUpdate){echoSuccess updated; } else {die('Invalid query:'.mysql_error()); }
?>
< a href =index.php>返回索引< / a>
任何帮助将不胜感激。这件事让我疯狂。
不同的问题:见上面的update.php代码。
当我点击名为更新信息的数据库条目旁边的链接时,我需要更新.php?id = charactersid
目前,它显示一个空的输入表单。我可以输入信息并点击更新,它会正确更新。
然而,当我被带到update.php页面时,我希望它显示输入除了值之外的格式应该等于数据库中已有的值。
例如:
目前它是什么显示:
名称:[空输入框]
我希望它显示:
名称:[在数据库中列出的当前名称]
所以输入的值应该是该字符的信息;所以当我想要更新时,我可以看到已经列出的那个角色的信息,并且改变/删除/添加任何我需要的东西。
更有意义吗?
你正以相反的方式使用它:
string mysqli_real_escape_string(mysqli $ link,string $ escapestr)
所以它应该是:
$ p $ $ name = mysqli_real_escape_string($ con,trim($ _ POST [name]));
来源: http://php.net/mysqli_real_escape_string
由于您使用的是MySQLi,我建议您跳入准备好的语句而不是real_escape,像这样:
<?php
//您的数据库信息
$ db_host ='host' ;
$ db_user ='user';
$ db_pass ='pass';
$ db_name ='database';
// POST数据
$ id = $ _POST ['id'];
$ name = trim($ _ POST [name]);
$ con = mysqli_connect($ db_host,$ db_user,$ db_pass,$ db_name);
if($ con> connect_error)
{
die('Connect Error('。mysqli_connect_errno()。')'。mysqli_connect_error());
}
$ sql =UPDATE characters SET name =?WHERE id =?;
if(!$ result = $ con> prepare($ sql))
{
die('Query failed:('。$ con> errno。')'。 $ CON组>错误);
$ b $ if(!$ result-> bind_param('si',$ name,$ id))
{
die('绑定参数失败: ('。$ result-> errno。')'。$ result-> error);
}
if(!$ result-> execute())
{
die('执行失败:('。$ result-> errno。 ')'。$ result->错误);
}
$ result-> close();
$ con> close();
echo成功更新;
?>
< a href =index.php>返回索引< / a>
选择字符名称:
<?php
//您的数据库信息
$ db_host ='host';
$ db_user ='user';
$ db_pass ='pass';
$ db_name ='database';
// POST数据
$ id = $ _POST ['id'];
$ con = mysqli_connect($ db_host,$ db_user,$ db_pass,$ db_name);
if($ con> connect_error)
{
die('Connect Error('。mysqli_connect_errno()。')'。mysqli_connect_error());
}
$ sql =SELECT name FROM characters WHERE id =?;
if(!$ result = $ con> prepare($ sql))
{
die('Query failed:('。$ con> errno。')'。 $ CON组>错误);
$ b $ if(!$ result-> bind_param('i',$ id))
{
die('绑定参数失败:('。 $ result-> errno。')'。$ result-> error);
}
if(!$ result-> execute())
{
die('执行失败:('。$ result-> errno。 ')'。$ result->错误);
}
$ result-> store_result();
if($ result-> num_rows == 0)
{
die('找不到字符......');
}
$ result-> bind_result($ name);
$ result-> fetch();
$ result-> close();
$ con> close();
echo $ name;
I have a secured area of my site where I can add/update/delete database entries. I am trying to update the script from mysql to mysqli. Everything works except for the "update" part. When I fill in the new information and click "update", it gives me this error:
Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, string given in /home3/tarb89/public_html/aususrpg.net/charbase/updated.php on line 19
I'm not sure what is wrong here; I'm a complete newbie, so my apologies.
Here is my update.php code:
<?php
// Create connection
$con=mysqli_connect("xxx","xxx","xxx","xxx");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$id = $_GET['id'];
$result = mysqli_query($con,"SELECT * FROM characters WHERE id = '$id'");
$my_array = array($c_z);
extract($my_array);
?>
<form id="FormName" action="../charbase/updated.php" method="post" name="FormName">
<table width="448" border="0" cellspacing="2" cellpadding="0">
<tr>
<td width="150" align="right"><label for="name">Name: </label></td>
<td><input name="name" maxlength="255" type="text" value="<?php echo stripslashes($name) ?>"></td>
</tr>
<tr>
<td colspan="2" align="center">
<input name="" type="submit" value="Update">
<input name="id" type="hidden" value="<?php echo $id ?>">
</td>
</tr>
</table>
</form>
The other issue I have with the update.php page is when it displays the table, the inputs should display the information already listed in the database; currently, it shows the name field but the input section is empty. It should display the name data already in the database (then I can replace it with whatever I want it to update as.)
Here is the updated.php code:
<?php
// Create connection
$con=mysqli_connect("xxx","xxx","xxx","xxx");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$id = $_POST['id'];
$name = mysqli_real_escape_string(trim($_POST["name"]), $con);
$rsUpdate = mysqli_query($con,"UPDATE characters SET name='$name'
WHERE id='$id'");
if($rsUpdate) { echo "Successfully updated"; } else { die('Invalid query: '.mysql_error()); }
?>
<a href="index.php">Back to index</a>
Any help would be greatly appreciated. This thing is driving me bonkers.
Different issue: see above update.php code.
When I click on a link next to a database entry, called "Update Information", it takes me to update.php?id=charactersid
Currently, it displays an empty input form. I can input information and hit "update" and it WILL update correctly.
However, when I am taken to the update.php page, I want it to display the input form except the values should all equal what is already in the database.
For example:
What it currently shows:
Name: [empty input box]
What I want it to show:
Name: [current name listed in the database]
So the value of the input should be the information for that character; so that when I want to update, I can see what is already listed as information for that character and change/delete/add whatever else I need to.
Does that make more sense?
You're using it the opposite way:
string mysqli_real_escape_string ( mysqli $link , string $escapestr )
So it should be:
$name = mysqli_real_escape_string($con, trim($_POST["name"]));
Source: http://php.net/mysqli_real_escape_string
Since you're using MySQLi I would suggest you to just jump into prepared statements rather than real_escape, like this:
<?php
// Your database info
$db_host = 'host';
$db_user = 'user';
$db_pass = 'pass';
$db_name = 'database';
// POST data
$id = $_POST['id'];
$name = trim($_POST["name"]);
$con = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if ($con->connect_error)
{
die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());
}
$sql = "UPDATE characters SET name = ? WHERE id = ?";
if (!$result = $con->prepare($sql))
{
die('Query failed: (' . $con->errno . ') ' . $con->error);
}
if (!$result->bind_param('si', $name, $id))
{
die('Binding parameters failed: (' . $result->errno . ') ' . $result->error);
}
if (!$result->execute())
{
die('Execute failed: (' . $result->errno . ') ' . $result->error);
}
$result->close();
$con->close();
echo "Successfully updated";
?>
<a href="index.php">Back to index</a>
To select the character name:
<?php
// Your database info
$db_host = 'host';
$db_user = 'user';
$db_pass = 'pass';
$db_name = 'database';
// POST data
$id = $_POST['id'];
$con = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if ($con->connect_error)
{
die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());
}
$sql = "SELECT name FROM characters WHERE id = ?";
if (!$result = $con->prepare($sql))
{
die('Query failed: (' . $con->errno . ') ' . $con->error);
}
if (!$result->bind_param('i', $id))
{
die('Binding parameters failed: (' . $result->errno . ') ' . $result->error);
}
if (!$result->execute())
{
die('Execute failed: (' . $result->errno . ') ' . $result->error);
}
$result->store_result();
if ($result->num_rows == 0)
{
die('No character found...');
}
$result->bind_result($name);
$result->fetch();
$result->close();
$con->close();
echo $name;
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