PHP，错误 1136：列数与第 1 行的值数不匹配
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PHP, Error 1136 : Column count doesn't match value count at row 1
提问by user701565
I get this Exception:
我得到这个异常：
Error 1136 : Column count doesn't match value count at row 1
Structure of the table :
表的结构：
create table gb_entries (
id int(4) not null auto_increment,
username varchar(40) not null,
name varchar(40),
gender varchar(40),
dob int(40),
email varchar(40),
primary key (id)
);
With this PHP code:
使用此 PHP 代码：
// Add a new entry to the database
function addEntry($username, $name, $gender, $dob, $email) {
$connection = mysql_open();
$insert = "insert into gb_entries " .
"values ('$username', '$name', '$gender', '$dob', '$email')";
$result = @ mysql_query ($insert, $connection)
or showerror();
mysql_close($connection)
or showerror();
}
// Return an array of database entries that contain $name anad $email
function getEntries($username,$name,$gender,$dob,$email) {
// Sanitise user input to prevent SQL injection attacks
$username = mysql_escape_string($username);
$name = mysql_escape_string($name);
$gender = mysql_escape_string($gender);
$dob = mysql_escape_string($dob);
$email = mysql_escape_string($email);
// Open connection and select database
$connection = mysql_open();
// Construct query
$query =
"select username, name, gender, dob, email from gb_entries where 0=0 ";
if (! empty($username)) {
$query .= "AND username LIKE '%$username%' ";
}
if (! empty($name)) {
$query .= "AND name LIKE '%$name%' ";
}
if (! empty($gender)) {
$query .= "AND gender LIKE '%$gender%' ";
}
if (! empty($dob)) {
$query .= "AND dob LIKE '%$dob%' ";
}
if (! empty($email)) {
$query .= "AND email LIKE '%$email%' ";
}
$query .= "ORDER BY id";
// echo $query;
// Execute query
$result = @ mysql_query($query, $connection)
or showerror();
// Transform the result set to an array (for Smarty)
$entries = array();
while ($row = mysql_fetch_array($result)) {
$entries[] = $row;
}
mysql_close($connection)
or showerror();
return $entries;
}
What does the Exception mean?
异常是什么意思？
回答by AgentConundrum
As it says, the column count doesn't match the value count. You're providing five values on a six column table. Since you're not providing a value for id, as it's auto increment, it errors out - you need to specify the specific columns you're inserting into:
正如它所说，列数与值数不匹配。您在六列表上提供五个值。由于您没有为 提供值id，因为它是自动递增的，因此会出错 - 您需要指定要插入的特定列：
$insert = "insert into gb_entries (username, name, gender, dob, email) " .
"values ('$username', '$name', '$gender', '$dob', '$email')"
Also, I really hate that WHERE 0=0line. I know whyyou're doing it that way, but I personally find it cleaner to do something like this (warning: air code!):
另外，我真的很讨厌那条WHERE 0=0线。我知道为什么你做这样的说法，但我个人觉得清洁剂做这样的事情（警告：空气代码）：
$query = "select username, name, gender, dob, email from gb_entries ";
$where = array();
if (! empty($username)) {
$where[] = "username LIKE '%$username%'"; // add each condition to an array
// repeat for other conditions
// create WHERE clause by combining where clauses,
// adding ' AND ' between conditions,
// and append this to the query if there are any conditions
if (count($where) > 0) {
$query .= "WHERE " . implode($where, " AND ");
}
This is personal preference, as the query optimizer would surely strip out the 0=0on it's own and so it wouldn't have a performance impact, but I just like my SQL to have as few hacks as possible.
这是个人偏好，因为查询优化器肯定会0=0自行剥离它，因此不会对性能产生影响，但我只是希望我的 SQL 尽可能少地进行黑客攻击。
回答by Pascal MARTIN
If the error is occurring when trying to insert a row to your table, try specifying the list of fields, in the insert query -- this way, the number of data in the valuesclause will match the number of expected columns.
如果在尝试向表中插入行时发生错误，请尝试在插入查询中指定字段列表——这样，values子句中的数据数将与预期的列数相匹配。
Else, MySQL expects six columns : it expects the idcolumn -- for which you didn't specify a value.
否则，MySQL 需要 6 列：它需要id您未指定值的列。
Basically, instead of this :
基本上，而不是这个：
$insert = "insert into gb_entries " .
"values ('$username', '$name', '$gender', '$dob', '$email')";
Use something like that :
使用类似的东西：
$insert = "insert into gb_entries (username, name, gender, dob, email) " .
"values ('$username', '$name', '$gender', '$dob', '$email')";
回答by pkanane
I had a similar problem. The column count was correct. the problem was that i was trying to save a String (the value had quotes around it) in an INT field. So your problem is probably coming from the single quotes you have around the '$dob'. I know, the mysql error generated doesn't make sense..
我有一个类似的问题。列数是正确的。问题是我试图在 INT 字段中保存一个字符串（该值周围有引号）。所以你的问题可能来自你在'$dob'周围的单引号。我知道，生成的 mysql 错误没有意义..
funny thing, I had the same problem again.. and found my own answer here (quite embarrassingly) It's an UNEXPECTED Data problem (sounds like better error msg to me). I really think, that error message should be looked at again
有趣的是，我又遇到了同样的问题......并在这里找到了我自己的答案（非常尴尬）这是一个意外的数据问题（对我来说听起来像是更好的错误信息）。我真的认为，应该再次查看该错误消息
回答by Chris
Does modifying this line help?
修改这条线有帮助吗？
$insert = "insert into gb_entries (username, name, gender, dob, email) " .
"values ('$username', '$name', '$gender', '$dob', '$email')";
