$conf, $runtime; function_exists('chdir') AND chdir(APP_PATH); $r = 'mysql' == $conf['cache']['type'] ? website_set('runtime', $runtime) : cache_set('runtime', $runtime); } function runtime_truncate() { global $conf; 'mysql' == $conf['cache']['type'] ? website_set('runtime', '') : cache_delete('runtime'); } register_shutdown_function('runtime_save'); ?>javascript - PHP: onClick not working - Stack Overflow|Programmer puzzle solving
最新消息:Welcome to the puzzle paradise for programmers! Here, a well-designed puzzle awaits you. From code logic puzzles to algorithmic challenges, each level is closely centered on the programmer's expertise and skills. Whether you're a novice programmer or an experienced tech guru, you'll find your own challenges on this site. In the process of solving puzzles, you can not only exercise your thinking skills, but also deepen your understanding and application of programming knowledge. Come to start this puzzle journey full of wisdom and challenges, with many programmers to compete with each other and show your programming wisdom! Translated with DeepL.com (free version)

javascript - PHP: onClick not working - Stack Overflow

matteradmin14PV0评论

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...

Share Improve this question edited Sep 7, 2013 at 14:06 ComFreek 29.5k18 gold badges107 silver badges157 bronze badges asked Sep 7, 2013 at 13:43 Princess ToledoPrincess Toledo 811 gold badge3 silver badges11 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
Add a ment  | 

2 Answers 2

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

Post a comment

comment list (0)

  1. No comments so far