PHP data modification problem, hope help!

You put $mysql=mysql query("select * from user where name='$name'");Change to $mysql=mysql query("select * from user where name='”.$name."'");Then print it in the update file to see if there is a value.

Also, put the data update in front of the query in the index.

You insert data from less to more like this.

query="update xianchang set year='$year'where id=$id";And so little by little.

Then write an echo $query and put the statement directly into phpmyadmin to test, but it seems that your main problem is that you wrote it incorrectly.

query="update ..

query=mysql_query($query_update);Yours should be $query update="update ..

That's a bunch.

$query=mysql_query($query_update);

Here it should be $query=mysql query($query);

It can be modified to let him report the error so that it is easier to determine the cause of the error $query1=mysql query($query) or die(mysql error()).

I've tried this, and if your post parameter is fine, and the database field is fine, you should be fine. Prefer $query rather than $query update

You submit with a post submission should be fetched with $post.

t7=$_post[t7];

What does get param do with the data? A little interested, hehe.

You can't update what you said, at least you should post the error message too, so that others can help you.

You get the param() function, all of which are $ post, and then output it, and see if you get it.

Then update a message first, such as update xianchang set year='$_post[year]' where id=$id

If the update is not successful, you can take a look at the SQL statement output, or copy the SQL statement to phpmyadmin to confirm that there is no problem.

If it doesn't work anymore, maybe mysql strikes.

You don't need to modify so much data at the same time, but make a few changes to see which part is wrong.

Step 1: On the computer Win7 system, press the windows key + R and enter cmd [Enter].

Step 4: set password = passwrod('New passwords');

This skill is very important!!

You're missing the update template.

You go and check whether the ID in the update template is passed in the get mode or post.

For example. This id is the get method.

This is the POST method.

Your two lines of ** are written in question, and it should be.

Also, your update statement should have where id=".get["id"]；

The form values passed through the post method of the form are obtained with the post, and the values passed by the browser address are obtained with the get method.

Also, your ** is very non-normal, and the entries in $get and $post are best enclosed in double quotation marks.

Your mistake is mainly here.

sql="update symbols set country = post[country]',animal = post[animal]',cname='$post[cname_]'where ";

When Chinese characters are present, Chinese characters are used''Bracket, and English characters are not used, otherwise it will be wrong.

Like this, full stop. If you can omit the table name in a single table, you can use the id, and only in a multi-table query do you need to indicate which form a field belongs to.

and your $result=$mysqli->query(sql) below; SQL should be preceded by a $.

Here's what I changed**, which has been tested and can be used conveniently:

One of the weird things is that $ post[cname], which is automatically added as cname in the post form, is the result of the var dump ($ post) test, and the var dump function is used to see the variable type and variable value. There is also the change of the form to the later.

echo ""

echo "

result->close();

where id='1''Wrong, pay attention to the quotation marks.

Or not'$post[telephone], there is a problem with the quotation marks anyway.

Note that the single quotation marks of the SQL statement corresponding to the field type are used.

Try putting now() in single gravity.