[英]Fatal error: Call to a member function bind_param() on a non-object
I keep getting this error我不断收到此错误
Fatal error: Call to a member function bind_param() on a non-object in H:\\USBWebserver\\root\\GIPSite\\stagetoevoegen.php on line 46致命错误:在第 46 行的 H:\\USBWebserver\\root\\GIPsite\\stagetoevoegen.php 中的非对象上调用成员函数 bind_param()
After a lot of research and tried many possible solutions, it still gives the error Here is the code经过大量研究并尝试了许多可能的解决方案,它仍然给出错误这是代码
if (isset($_POST['cmdVerstuur']))
{
$invoegen=$link->prepare("insert into tblstagebedrijven (Naam,Begeleider, Locatie, Telefoon, E-mail, Opmerking) values(?,?,?,?,?,?)");
$naam = $_POST['txtNaambedrijf'];
$begeleider = $_POST['txtBegeleider'];
$locatie = $_POST['txtLocatie'];
$telefoon = $_POST['txtTelefoon'];
$email = $_POST['txtEmail'];
$opmerking = $_POST['txtOpmerking'];
$invoegen->bind_param('ssssss', $naam, $begeleider, $locatie, $telefoon, $email, $opmerking);
$resultaat=$invoegen->execute();
if($resultaat)
echo "Het stagebedrijf is toegevoegd";
else
echo "Er is een fout opgetreden";
$link->close();
}
I have a connection.php file that is included and other pages do work with it.我有一个包含在内的 connection.php 文件,其他页面可以使用它。
Edit: I tested your code and was successful ( in conjunction with my answer ).编辑:我测试了您的代码并成功(结合我的回答)。
See my troubleshooting-debugging notes below.请参阅下面的故障排除调试说明。
Taking out the backticks around the E-mail
column gave the following error message similar to this and using error reporting such as:去掉E-mail
列周围的反引号给出了与此类似的以下错误消息,并使用错误报告,例如:
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
at the top under the opening <?php
tag在开头<?php
标签下的顶部
Fatal error: Uncaught exception 'mysqli_sql_exception' with message 'You have an error in your SQL syntax;致命错误:未捕获的异常“mysqli_sql_exception”,消息为“您的 SQL 语法有错误; check the manual that corresponds to your MySQL server version for the right syntax to use near '-mail, Opmerking) values(?,?,?,?,?,?)' at line 1' in /home/user/public_html/dbtest1.php:17 Stack trace: #0 /home/user/public_html/dbtest1.php(17): mysqli->prepare('insert into tbl...') #1 {main} thrown in /home/user/public_html/dbtest1.php on line 17检查与您的 MySQL 服务器版本相对应的手册,了解在 /home/user/public_html/ 中第 1 行的“-mail, Opmerking) values(?,?,?,?,?,?)”附近使用的正确语法dbtest1.php:17 堆栈跟踪:#0 /home/user/public_html/dbtest1.php(17): mysqli->prepare('insert into tbl...') #1 {main} 扔在 /home/user/第 17 行的 public_html/dbtest1.php
Wrap your E-mail
column value in backticks or choose another way, Email
or E_mail
用反引号包裹您E-mail
列值或选择另一种方式, Email
或E_mail
(Naam,Begeleider, Locatie, Telefoon, `E-mail`, Opmerking)
You can use punctuation, white space, international characters, and SQL reserved words if you use delimited identifiers.如果使用分隔标识符,则可以使用标点符号、空格、国际字符和 SQL 保留字。
I quote from this answer : (which I already knew)我引用了这个答案:(我已经知道了)
The major reason against use of hyphen is that most references must then quote the field names.反对使用连字符的主要原因是大多数引用必须引用字段名称。 Otherwise they will look like a subtraction operator, both to MySQL and humans.否则,对于 MySQL 和人类来说,它们将看起来像一个减法运算符。
Using error reporting would have caught that.使用错误报告会发现这一点。
Example:例子:
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
Error checking methods (MySQL-PDO)错误检查方法(MySQL-PDO)
MySQL MySQL
// connect (create a new MySQLi object) with custom predefined constants
$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_TABLE_NAME);
// $mysqli->connect_error is buggy until PHP 5.3.0
if (mysqli_connect_error())
{
throw new Exception(mysqli_connect_error());
}
$result = $mysqli->query($sql);
if (!$result)
{
throw new Exception($mysqli->error);
}
while($row = $mysqli->fetch_assoc($result))
{
// your result handling code (print it)
}
PDO PDO
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
Footnotes: (troubleshooting-debugging)脚注:(故障排除-调试)
Place var_dump($link->error);
放置var_dump($link->error);
before $invoegen->bind_param('ssssss'
在$invoegen->bind_param('ssssss'
a) You can also try this method: a) 你也可以试试这个方法:
$invoegen = $mysqli->prepare("...");
if( $invoegen !== FALSE ) {
$invoegen->bind_param(...);
$invoegen->execute();
}
http://php.net/manual/en/mysqli.prepare.php http://php.net/manual/en/mysqli.prepare.php
b) Check to see what your column types are. b) 检查您的列类型。
c) Place var_dump($invoegen);
c) 放置var_dump($invoegen);
right after your prepare call.在您的准备电话之后。
d) Check your column names for spelling and/or actual existence. d) 检查您的列名称的拼写和/或实际存在。
e) Check to see if your form elements are named and with no typos, checking for letter-case also. e) 检查您的表单元素是否已命名且没有拼写错误,还要检查字母大小写。 A!=a;
meaning that an uppercase A
is not treated the same as a
.意味着大写A
与a
。 For example: name="Animal"
as opposed to name="animal"
those are two different animals altogether.例如: name="Animal"
与name="animal"
相反,它们完全是两种不同的动物。
The code I used to test it with: - All columns set to VARCHAR(255)
我用来测试它的代码: - 所有列都设置为VARCHAR(255)
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$DB_HOST = "xxx";
$DB_NAME = "xxx";
$DB_PASS = "xxx";
$DB_USER = "xxx";
$link = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($link->connect_errno > 0) {
die('Connection failed [' . $link->connect_error . ']');
}
$invoegen=$link->prepare("insert into tblstagebedrijven (Naam,Begeleider, Locatie, Telefoon, `E-mail`, Opmerking) values(?,?,?,?,?,?)");
/*
$naam = $_POST['txtNaambedrijf'];
$begeleider = $_POST['txtBegeleider'];
$locatie = $_POST['txtLocatie'];
$telefoon = $_POST['txtTelefoon'];
$email = $_POST['txtEmail'];
$opmerking = $_POST['txtOpmerking'];
*/
$naam = "Fred";
$begeleider = "txtBegeleider";
$locatie = "txtLocatie";
$telefoon = "txtTelefoon";
$email = "txtEmail";
$opmerking = "txtOpmerking";
$invoegen->bind_param('ssssss', $naam, $begeleider, $locatie, $telefoon, $email, $opmerking);
$resultaat=$invoegen->execute();
if($resultaat)
echo "Het stagebedrijf is toegevoegd - OK";
else
echo "Er is een fout opgetreden - NOT ok";
$link->close();
... n, E-mail, Opmerking) v...
^---- seen as a mathematical subtraction operation
You need to quote it:你需要引用它:
... n, `E-mail`, Opmerking) v...
Plus, if you'd had proper error handling in your code, you'd have been told that your prepare operation failed and returned a boolean false:另外,如果您在代码中进行了正确的错误处理,您会被告知准备操作失败并返回布尔值 false:
$stmt = $link->prepare(...);
if (!$stmt) {
die($link->errorInfo); // or whatever your DB lib's error reporting method is
}
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