PHP：onClick不工作

Question

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.org/image/wbiigechn/

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

Answer 1

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()">