警告：mysqli_query（）期望参数2是string，给定的对象 [英] Warning: mysqli_query() expects parameter 2 to be string, object given in
问题描述
早上好！
我是PHP新手。我正在努力使这部分工作，但显示我这个问题。它是一个修改mysql数据库中的一些记录的形式。代码获取数据，但它显示我的错误，当我运行修改它显示我几个问题....
警告：mysqli_query （）期望参数2是字符串，在第15行的C：\wamp\www\CTE\formedicion.php中给出
的对象。
我仍然不知道如何解决这个问题。
谢谢！
< body> ;
<？php
包含conexiondb.php;
如果（！isset（$ _ POST ['submit']））{
// $ busqueda = $ con> query（
$ muestra = $ con> query（SELECT * FROM clientes C INNER JOIN producto P ON C.serial = P.serial WHERE P.serial = $ _GET [serial]）;
// mysqli_query（ $ con $，$ sql）;
mysqli_query（$ con，$ muestra）;
$ person = $ muestra-> fetch_array（）;
}
？ >
< form action =<？php echo $ _SERVER ['PHP_SELF'];？>method =post>
cliente< input type =textname =inputclientevalue =<？php echo $ person ['cliente'];？>/>< br />
cedula< ; input type =textname =inputcedulavalue =<？php echo $ person ['cedula'];？>/>< br />
< ; input type =hiddenname =serialvalue =<？php echo $ _GET ['serial'];？>/>
< input type =submitname =提交value =Modificar/>
< / form>
<？php
if（isset（$ _ POST ['submit']））{
$ u =更新E cliente SET'cliente'='$ _ POST [inputcliente]'，'cedula'='$ _POST [inputcedula]'WHERE serial = $ _ POST [serial];
mysqli_query（$ con，$ u）;
echoEl usuario ha sido modificado;
header（Location：busca.php）;
} else {
}
？>
< / body>
< / html>
更改
$ muestra = $ con> query（SELECT * FROM clientes C
INNER JOIN producto P ON C.serial = P.serial WHERE P.serial = $ _GET [serial]）;
至
//更新为清洁输入
$ serial = mysqli_real_escape_string（$ con，$ _ GET [serial]）;
$ qry =SELECT * FROM clientes C
INNER JOIN producto P ON C.serial = P.serial WHERE P.serial = $ serial;
然后
$ muestra = mysqli_query $ con，$ qry）;
Good morning!
I'm new in PHP. i'm trying to make work this scrpt but is showing me this problem. It's a forma that modifies some records in a mysql database. The codes gets the data but it shows me that mistake and when i run modify it shows me several problems....
Warning: mysqli_query() expects parameter 2 to be string, object given in C:\wamp\www\CTE\formedicion.php on line 15
I still don't know how to fix it. I really appreaciate your help.
Thanks!
<body>
<?php
include "conexiondb.php";
if(!isset($_POST['submit'])){
//$busqueda=$con->query(
$muestra=$con->query("SELECT * FROM clientes C INNER JOIN producto P ON C.serial = P.serial WHERE P.serial = $_GET[serial]");
//mysqli_query($con,$sql);
mysqli_query($con,$muestra);
$person=$muestra->fetch_array();
}
?>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
cliente<input type = "text" name="inputcliente" value = "<?php echo $person['cliente']; ?>" /><br/>
cedula <input type = "text" name="inputcedula" value = "<?php echo $person['cedula']; ?>" /><br/>
<input type="hidden" name="serial" value="<?php echo $_GET['serial'];?>"/>
<input type = "submit" name = "submit" value= "Modificar"/>
</form>
<?php
if (isset($_POST ['submit'])){
$u = "UPDATE cliente SET'cliente'='$_POST[inputcliente]','cedula' = '$_POST[inputcedula]' WHERE serial=$_POST[serial]";
mysqli_query($con,$u);
echo "El usuario ha sido modificado";
header ("Location:busca.php");
} else {
}
?>
</body>
</html>
Change
$muestra=$con->query("SELECT * FROM clientes C
INNER JOIN producto P ON C.serial = P.serial WHERE P.serial = $_GET[serial]");
To
// Updated to clean the input
$serial = mysqli_real_escape_string($con,$_GET[serial]) ;
$qry="SELECT * FROM clientes C
INNER JOIN producto P ON C.serial = P.serial WHERE P.serial = $serial";
Then
$muestra = mysqli_query($con,$qry);
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