javascript - PHP: onClick not working - Stack Overflow

My button "VIEW" is not going to the form of "booking_content.php" instead its just refreshing the page of "home.php"

Here's my code:

$radio = mysql_query("SELECT fldBldgName, MAX(fldTotalDuration) as fldTotalDuration FROM tbldata WHERE fldNetname = '".$get_radio."' AND fldMonth = '".$get_month."' AND fldWeek = '".$get_week. "' GROUP BY fldBldgName ORDER BY id, fldBldgName, fldTotalDuration DESC");

echo "<table class = 'tblMain'>";
echo "<tr align='left'>";
echo "<td><b><u>BUILDING NAME</u></b></td>";
while ($row = mysql_fetch_array($radio))
{   
echo "<tr><td align='left'>";   
echo $row['fldBldgName']."'>";

echo "<input type='image'  src='image/view.png' name='viewBldg' onClick='this.form.action='booking_content.php'; this.form.submit()'>";
echo $row['fldBldgName'];
}
echo "</tr></table>";

My problem is this one:

  echo "<input type='image'  src='image/view.png' name='viewBldg' onClick='this.form.action='booking_content.php'; this.form.submit()'>";   

*The onClick is not going to the page of booking_content.php instead its just refreshing the page...

I can't upload an example result of my program...

Please click this link so you can view my sample program: /

ALSO the code for my button is inside of a fieldset...

My button "VIEW" is not going to the form of "booking_content.php" instead its just refreshing the page of "home.php"

Here's my code:

$radio = mysql_query("SELECT fldBldgName, MAX(fldTotalDuration) as fldTotalDuration FROM tbldata WHERE fldNetname = '".$get_radio."' AND fldMonth = '".$get_month."' AND fldWeek = '".$get_week. "' GROUP BY fldBldgName ORDER BY id, fldBldgName, fldTotalDuration DESC");

echo "<table class = 'tblMain'>";
echo "<tr align='left'>";
echo "<td><b><u>BUILDING NAME</u></b></td>";
while ($row = mysql_fetch_array($radio))
{   
echo "<tr><td align='left'>";   
echo $row['fldBldgName']."'>";

echo "<input type='image'  src='image/view.png' name='viewBldg' onClick='this.form.action='booking_content.php'; this.form.submit()'>";
echo $row['fldBldgName'];
}
echo "</tr></table>";

My problem is this one:

  echo "<input type='image'  src='image/view.png' name='viewBldg' onClick='this.form.action='booking_content.php'; this.form.submit()'>";   

*The onClick is not going to the page of booking_content.php instead its just refreshing the page...

I can't upload an example result of my program...

Please click this link so you can view my sample program: http://postimg/image/wbiigechn/

ALSO the code for my button is inside of a fieldset...

• php
• javascript
• jquery
• button
• onclick

Share Improve this question Follow edited Sep 7, 2013 at 14:06 ComFreek 29.5k1818 gold badges107107 silver badges157157 bronze badges asked Sep 7, 2013 at 13:43 Princess ToledoPrincess Toledo 8111 gold badge33 silver badges1111 bronze badges 2

• Where is the form tag in your code? – SaidbakR Commented Sep 7, 2013 at 13:52
• @sємsєм..its in the upper part of my home.php...I didn't include it in the question...but heres the form... <form name='form' method='post' action="">, the code is inside of the form... – Princess Toledo Commented Sep 7, 2013 at 13:54

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2 Answers 2

Sorted by: Reset to default 3

Your attempt will generate the following HTML:

<input type='image' src='image/view.png' name='viewBldg' 
onclick='this.form.action='booking_content.php'; this.form.submit()'>

So your browser wont be able to interpret the onclick attribute correctly, because the ' before booking_content. Try the following:

echo "<input type='image' src='image/view.png' name='viewBldg' 
onclick=\"this.form.action='booking_content.php'; this.form.submit()\">";

Which should generate a valid HTML such as:

<input type='image' src='image/view.png' name='viewBldg' 
onclick="this.form.action='booking_content.php'; this.form.submit()">

You tried to access your form in wrong way. this in your code refer to the input itself not to the form.

You may have to do something like the following:

<input type="button" value="view" onclick="document.fn.action='http://google.';document.fn.submit();" />

where fn the attribute name value of your form.

Notice: in this case you should have only one form with that name.

Check out this demo: http://jsbin./asekAqu/1 or its sample code: http://jsbin./asekAqu/1/edit