parse errors no longer visible View as Markdown

I have no problem getting a parse error to be shown, can you describe what
you did to get that behavior?

Zeev

At 16:53 23/04/2003, Derick Rethans wrote:

Hello,

I was trying to make a screenshot of a parse error today, but found out
that I was unable to get it shown on the screen. Error reporting was set
to 2047 in php.ini, but still nothing to see (only in my apache error
log). So I tracked the calls into php_error_cb, and see here the
results:

(gdb) run -X
Starting program: /dat/dev/php/apache_1.3.27/src/httpd -X

Program received signal SIGINT, Interrupt.
0x407c15be in __select () at __select:-1
-1 __select: No such file or directory.
in __select
(gdb) bre php_error_cb
Breakpoint 1 at 0x80aefce: file /dat/dev/php/php-5.0.0dev/main/main.c,
line 584.(gdb) cont
Continuing.

Breakpoint 1, php_error_cb (type=4,
error_filename=0x841ae84 "/home/httpd/html/test/xdebug_error.php",
error_lineno=3, format=0x84d1710 "parse error", args=0xbfffbbe8)
at /dat/dev/php/php-5.0.0dev/main/main.c:584
warning: Source file is more recent than executable.

584 buffer_len = vspprintf(&buffer, PG(log_errors_max_len),
format, args);
(gdb) next
585 if (PG(ignore_repeated_errors)) {
(gdb) next
595 display = 1;
(gdb) next
597 strlcpy(last_error.buf, buffer, sizeof(last_error.buf));
(gdb) next
598 strlcpy(last_error.filename, error_filename,
sizeof(last_error.filename));
(gdb) next
599 last_error.lineno = error_lineno;
(gdb) next
602 if (display && (EG(error_reporting) & type || (type & E_CORE))
(gdb) print executor_globals.error_reporting
$1 = 0

As you see the error_reporting level is '0' here, and thus nothing is
shown. I couldn't find why it is '0' here, but I suspect that the
following patch might be the problem:

http://news.php.net/article.php?group=php.zend-engine.cvs&article=638

Derick

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Derick Rethans http://derickrethans.nl/
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