Hi All
I'm having problems printing a string with a % char. Here is the code
line = sprintf("<table width=100% id=%s>\n", $table_id);
I'm getting an not enough parameter error. The % for the 100% is being
mis interepreted as a format char. Though i worked around by coding it
line = sprintf("<table width=100%s id=%s>\n", '%', $table_id);, i think
there should be a way with which the % can be escaped PHP.
The PHP documention does not have any info about this. if anyone has
come accross such problems and found a work around, please help me out.
Thanks
/V 6  2667 
Double % sign.
sprintf("<table width='100%%' id='%s'>\n", $table_id);
-Jay
Venkat Venkataraju wrote: Hi All
I'm having problems printing a string with a % char. Here is the code
line = sprintf("<table width=100% id=%s>\n", $table_id);
I'm getting an not enough parameter error. The % for the 100% is being
mis interepreted as a format char. Though i worked around by coding it
line = sprintf("<table width=100%s id=%s>\n", '%', $table_id);, i think
there should be a way with which the % can be escaped PHP.
The PHP documention does not have any info about this. if anyone has
come accross such problems and found a work around, please help me out.
Thanks
/V
Jay Moore wrote: Double % sign.
sprintf("<table width='100%%' id='%s'>\n", $table_id);
doesn't seems to work. it still asks for more parms. i also did \%, but
it is not working either.
/V
Venkat Venkataraju wrote:
Jay Moore wrote:
Double % sign.
sprintf("<table width='100%%' id='%s'>\n", $table_id);
doesn't seems to work. it still asks for more parms. i also did \%, but
it is not working either.
/V
Works just fine for me. Make sure you don't have any typos.
-Jay
Venkat Venkataraju wrote: Hi All
I'm having problems printing a string with a % char. Here is the code
line = sprintf("<table width=100% id=%s>\n", $table_id);
I'm getting an not enough parameter error. The % for the 100% is being
mis interepreted as a format char. Though i worked around by coding it
line = sprintf("<table width=100%s id=%s>\n", '%', $table_id);, i think
there should be a way with which the % can be escaped PHP.
The PHP documention does not have any info about this. if anyone has
come accross such problems and found a work around, please help me out.
Thanks
/V
If it's anything like C++ (as PHP usually is), and if I remember my c++
correctly, it's going to be %%. Try that.
-- Inuyasha4Life
In message <h1************ *********@newss vr28.news.prodi gy.com>, Venkat
Venkataraju <segfault@?.?.i nvalid> writes Hi All
I'm having problems printing a string with a % char. Here is the code
line = sprintf("<table width=100% id=%s>\n", $table_id);
I'm getting an not enough parameter error. The % for the 100% is being
mis interepreted as a format char. Though i worked around by coding it
line = sprintf("<table width=100%s id=%s>\n", '%', $table_id);, i think
there should be a way with which the % can be escaped PHP.
The PHP documention does not have any info about this. if anyone has
come accross such problems and found a work around, please help me
out.
You have a percent literal in the line. You can just concatenate it:
line = "<table width=100% id=" . $tableid . ">\n";
--
Five Cats
Email to: cats_spam at uk2 dot net
Venkat Venkataraju <segfault\"@\"n ospam.sbcglobal .net> wrote: line = sprintf("<table width=100% id=%s>\n", $table_id);
This will work:
line = sprintf("<table width=100%s id=%s>\n", '%', $table_id);
--
Spam is spam no matter who's doing it or for what reason. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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