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Javascript inside php function


souciance

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Hello,

 

I have a php function which prints out some html form code. I am having trouble getting the javascript to work when it is being echoed in the function. This is part of the code where the javascript is called.

 

 echo'<table class="currencyTransferTable" cellspacing="10" cellpadding="10">
		  <tr>  
	<td>		
	<label for="amount" class="label">Amount: </label>
	</td>
	<td>
	<input id="amount" name="amount" value="' .$amount;

	echo '" class="input" type="text" /><img src="images/asterisk.gif" height="7" class="img" alt="required" onkeypress="javascript:return numbersonly(event)" /><br>';		
	echo '</td></tr>';
	global $error;		

 

The problem is, not matter what javascript I write it fails to register. Even a simple alert() won't work. Are there special things to remember when printing out javascript, especially inside a php function?

Thanks, Moeed.

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Well from what I remember, you can't write javascript inside PHP

Of course you can! You can write anything that would be a complete page to be received by the browser: HTML, CSS, JavaScript, etc.

 

Have you checked the rendered HTML code to see if the JavaScript is generated as expected? There might be a quote mark out of place. The manner in which you are echoing the code makes it difficult to follow.

 

I do notice that you have a parameter in the onkeypress() trigger - event. has that variable been declared? Also, are you getting JavaScript errors?

 

here is a rewrite of your code that's a little easier to follow for me:

echo "<table class=\"currencyTransferTable\" cellspacing=\"10\" cellpadding=\"10\">\n";
echo "  <tr>\n";
echo "    <td><label for=\"amount\" class=\"label\">Amount: </label></td>\n";
echo "    <td>
echo "      <input id=\"amount\" name=\"amount\" value=\"$amount\" class=\"input\" type=\"text\" />\n"
echo "      <img src=\"images/asterisk.gif\" height=\"7\" class=\"img\" alt=\"required\" onkeypress=\"javascript:return numbersonly(event)\" /><br>\n";
echo "    </td>\n";
echo "  </tr>\n";

global $error;		

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