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Showing results for tags 'debugging'.
Found 6 results
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Dear all, I have the following code for storing an image in and fetching it from a MySQL DB: <html> <body> <form action="index.php" method="POST" enctype="multipart/form-data"> File: <input type="file" name="image"> <input type="submit" value="upload"> </form> <?php mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("code_new") or die(mysql_error()); $file = $_FILES['image']['tmp_name']; if(!isset($file)) { echo 'Please select a file'; } else { $image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
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Hi, This block code is from W3schools tutorials: <?php $cars = array("Volvo", "BMW", "Toyota"); $arrlength = count($cars); for($x = 0; $x <= $arrlength; $x++) { echo $cars[$x]; echo "<br>"; } ?> The original block was without the equal sign in $x <= $arrlength; I have added the equal sign to see what would happen. I got the same result but followed with: Notice: Undefined offset: 3 in C:\xampp\htdocs\Testing\testing.php on line 14 I would appreciate it if someone would explain why I am getting this notice and its meaning. I thought the loop will stop and giv
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Hi. I attempted to load a couple of PHP pages, one of which is the index page, but the message I got is below. The index page is also open in another tab; when refreshed it runs perfectly, but loading it afresh is deniable. Both Apache and MySQL are running; I turned them off and on again and ensured that skype is also off and no firewall set up is changed or updated. I would appreciate if someone could advise what could be the problem and what is the solution? Object not found! The requested URL was not found on this server. If you entered the URL manually please check your spelling and
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Dear all: I have the following php line of code: <?php $var2 = 5; ?> += : <?php $var2 += 4; echo var2; ?><br /> Because of the lack of the dollar sign before var2 following the echo, it spat out the following error: += : Notice: Use of undefined constant var2 - assumed 'var2' in C:xampphtdocsTutorialsPHPw3schoolsintegers.php on line 29var2 My understanding prior to this was that Notice means something bad in the code but it is not an error and would not stop the code from executing. But the code has not executed in this example. Since the code is not executing,
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How can I write my script to a web console (firefox) only when a create a boulean variable called debug set to equal true?
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I'm using a ZooEffect Gallery, which is great-except for two things; A) The images do not read links-despite them being a gallery option. You can't seem to change the font within the gallery-I tried being adjusting the CSS-to no avail, I put it back and then after a few days, ZooEffect got back to me with a solution, Which also didn't work. Most people seem to be able to pinpoint these issues using Safari Developer or FireBug. Can anybody pinpoint for me or at least tell me how I would do this? ThanksLauren
