On Sat, Mar 31, 2012 at 6:36 AM, Maciek Sokolewicz
<maciek.sokolewicz@xxxxxxxxx> wrote:
> On 31-03-2012 10:29, tamouse mailing lists wrote:
>>
>> On Sat, Mar 31, 2012 at 1:45 AM, saeed ahmed<mycomputerbooks@xxxxxxxxx>
>> wrote:
>>>
>>> i have made a php script with a tutorial help....i dont know where is a
>>> error.please have a look
>>
>>
>> Before you actually run code in a browser, check your syntax with php
>> -l (for "lint", the old command to check C syntax on unixy systems) on
>> a command line. This will improve your life immeasurably since you
>> won't have to chase down where the server logs errors just to see if
>> your code was syntactically correct.
>>
>> miishka:Sites tamara$ php -l test.saeed.php
>> PHP Parse error: syntax error, unexpected T_STRING in test.saeed.php on
>> line 6
>> Errors parsing test.saeed.php
>> miishka:Sites tamara$
>>
>> Which points out the syntax error on line 6 where you have:
>>
>> $result=mysql_query("select("SELECT*FROM COLLEAGUE");
>>
>> Which as an improperly formed select sql, and an unclosed string (thus
>> sucking up part of your php code and html document. To fix this, it
>> should be:
>>
>> $result=mysql_query($db, "SELECT * FROM COLLEAGUE");
>
>
> Wrong, it should be the other way around. First the query, and then an
> OPTIONAL link-identifier (which I assume $db to be in this case).
>
>>
>> i.e., the sql statement is in a single string. Also, spacing around
>> the column list "*" may or may not be necessary, but it's a lot easier
>> to read if it's there. Also, you need to tell the query procedure what
>> data base you're using the $db variable (see below for this).
>>
>> Ok, that fixed, run php -l on the file again, and we get:
>>
>> PHP Parse error: syntax error, unexpected '>' in test.saeed.php on
>> line 31
>> Errors parsing test.saeed.php
>>
>> Looking at line 31:
>>
>> echo "<td>" . $row['firstName'] ."</td>";
>>
>> it looks fine, you might be wondering why it's complaining about that
>> line. The key is to look up from there for possible syntax problems,
>> and on line 30, there is a problem in that you didn't close the last
>> string:
>>
>> echo"<td>".$row['id'] ."</td>;
>>
>> should be:
>>
>> echo"<td>".$row['id'] ."</td>";
>>
>>
>> Ok, fix that, run php -l again:
>>
>> PHP Parse error: syntax error, unexpected '[', expecting ',' or
>> ';' in test.saeed.php on line 34
>> Errors parsing test.saeed.php
>>
>> So, look at line 34 now:
>>
>> echo "<td>" . Srow['email'] ."</td";
>>
>> The problem is that you typoed the dollar sign in front of row. It should
>> be:
>>
>> echo "<td>" . $row['email'] ."</td";
>>
>> Also, on this line, although php -l won't catch it, is an html error.
>> The last /td is not closed, so:
>>
>> echo "<td>" . $row['email'] ."</td>";
>>
>> Run php -l again, and this time, voila:
>>
>> No syntax errors detected in test.saeed.php
>>
>> That solves your syntax errors. Now you can test it from a server.
>>
>> Before you do that though, a few other points:
>>
>> When developing code, it's a really good idea to turn on run time
>> error checking so you can see things in the browser when a problem
>> crops up:
>>
>> <?php
>> error_reporting(-1); // turns on EVERYTHING
>> ini_set('display_errors',true);
>> ini_set('display_startup_errors',true);
>>
>> // follow with your code...
>>
>>
>> This won't help if your code doesn't parse correctly, but that's what
>> the php -l is for above, but it will reveal run-time processing errors
>> in your browser. And you have several.
>>
>> As another responder said, checking the return values of functions is
>> important. Sometimes tutorials omit this for the sake of brevity, but
>> it is something that still should be done.
>>
>> Line 3:
>>
>> mysql_connect("localhost","root"," " );
>>
>> is incorrect on at least two levels. First, you haven't saved the data
>> base resource that is returned from mysql_connect, and second, you
>> haven't checked that the call was successful. Third, possibly is that
>> you have specified the root mysql password a string of one space. This
>> is unlikely to be the password.
>>
>> It should be rewritten as:
>>
>> $db = mysql_connect("localhost", "root", "") or die("Could not
>> connect to mysql server on local host: " . mysql_error() . PHP_EOL);
>>
>> The next line:
>>
>> mysql_select_db("addressbook");
>>
>> is also incorrect as it does not specify the data base resource which
>> you omitted in the previous call, and you aren't (again) checking the
>> result to see if it works.
>>
>> It should be rewritten as:
>>
>> mysql_select_db($db, "addressbook") or die ("Could not connect to
>> data base 'addressbook'. Error: " . mysql_error($db) . PHP_EOL);
>>
>> Back to that line 6, after you've repaired it, you *again* need to
>> check that the query actually worked. After that query, include the
>> following line:
>>
>> if (FALSE === $result) die ("Query failed. Error: " .
>> mysql_error($db) . PHP_EOL);
>>
>> Note that if the table is empty, that is not an error; your result set
>> will simply be empty. Returning FALSE indicates that there was an
>> actual error in the query. I find it better to put the sql statement
>> into a string, and then use the string variable in the actual query
>> function. This lets you also print out the query as it was sent to
>> mysql:
>>
>> $sql = "select * from colleague";
>> $result = mysql_query($db, $sql);
>
>
> Again, query first, linkidentifier after that. This way you'll always get an
> error.
>
>
>> if (FALSE === $result) die ("Query failed. \$sql=$sql. Error: " .
>> mysql_error($db) . PHP_EOL);
>>
>> A little further on, there is a problem in the html meta statement:
>> there is no closer on that line:
>>
>> <meta http-equiv="content-type" content="text/html;charset=utf-8"
>>
>> It needs to be closed with "/>" since you're using XHTML.
>>
>> Line 39 needs to have the data base resource as well that was set
>> above in the connect:
>>
>> mysql_close($db);
>>
>> You owe me one internet bheer for debugging your script. :)
>
>
> You made mistakes during debugging :s
>
Opps, so right!
I owe you an internet bheer.
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