JavaScript inline validation in php code

Question

Hi I want to validate a textfield with javascript. Here is my code in php:

$question_type = "<input type='text' id='question_type_n' onkeyup="allnumeric(this.value)" name='question_type_n_".$qid."' value='".$d[0]."' >";

I also have the function allnumeric() in javascript. But it returns errors. Javascript-Error:

Parse error: syntax error, unexpected 'allnumeric' (T_STRING) in E:\Share\xampp\htdocs\Dropbox\icsurvey\application\views\finish_survey.php on line 142

I think your problem is with nested quotes. Try single quotes instead.. $question_type = "<input type='text' id='question_type_n' onkeyup='allnumeric(this.value)' name='question_type_n_".$qid."' value='".$d[0]."' >"; — karthikr
– karthikr, Commented Sep 26, 2014 at 5:20

Mithun Satheesh · Accepted Answer · 2014-09-26 05:20:45Z

2

You have to escape it properly

$question_type = "<input type='text' id='question_type_n' onkeyup='allnumeric(this.value)' name='question_type_n_".$qid."' value='".$d[0]."' >";

First thing is that you dint concatenate the string properly. Second error being php wont understand the js function allnumeric. You have to embed it as string so that it will be triggered on the onclick event on your browser not at the serverside.

answered Sep 26, 2014 at 5:20

Mithun Satheesh

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Nikhil Musale · Accepted Answer · 2014-09-26 05:26:51Z

0

instead of using onkeyup="allnumeric(this.value)" use in single quotes like onkeyup='allnumeric(this.value)'

answered Sep 26, 2014 at 5:26

Nikhil Musale

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4 Comments

Olimpiu POP Over a year ago

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post.

Spencer Wieczorek Over a year ago

@user503413 This does provide a solution. Simple solutions are still answers.

Olimpiu POP Over a year ago

@SpencerWieczorek - thank you for your feedback, you could have used the time to edit the question to add some simple explanation why the provided code snippet would provide a solution and what was wrong in what the poster tried. And also, if you consider it that good as a response next to the question there is a up-vote button you can express your opinion in that way.

Spencer Wieczorek Over a year ago

@user503413 I'm simply stating that this is a valid answer, such that it provides a solution like the other answers. Not that it is a good answer, simply that it's not violating the rules. I don't think this is a good response, because the lack of explanation. And there is no need to edit because there are already better responses to this question.

Spencer Wieczorek · Accepted Answer · 2014-09-26 05:21:02Z

0

PHP outputs HTML, in which can contain JavaScript. Note you need to use . to concatenate in PHP. But in this case we just need to use single quotes:

$question_type = "<input ... onkeyup='allnumeric(this.value)' ... >";

answered Sep 26, 2014 at 5:21

Spencer Wieczorek

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Comments

jpcanamaque · Accepted Answer · 2014-09-26 05:21:10Z

0

Try this on your PHP code:

$question_type = "<input type='text' id='question_type_n' onkeyup='allnumeric(this.value)' name='question_type_n_".$qid."' value='".$d[0]."' >";

answered Sep 26, 2014 at 5:21

jpcanamaque

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