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Closed 7 years ago.
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I am unable to view images on my webserver in php and I keep getting server 500 error.
I believe that it is this line of code echo "<img src='$row["sourchPath"]'>":
while ($row = mysqli_fetch_assoc($result)){
if($row){
echo $row["sourchPath"]; // this works
echo "<img src='$row["sourchPath"]'>";
}
else {
echo "error";
}
}
My file directory is like images/football.jpg
The problem is because of this line:
echo "<img src='$row["sourchPath"]'>"; // syntax error
It should be,
echo "<img src='" . $row["sourchPath"] . "' />";
You may use
echo "<img src='" . $row["sourchPath"] . "'>";
in place of your code.
It will work
Related
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Closed 4 years ago.
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I have a simple link that needs to pass a variable through $_GET and set it to a variable in the page that is being opened. It is failing every time however and I am not sure why.
<a class='section_header' href='category/index.php?`id='" . $catid . "'><b>" . $row[0] ."</b></a>
Code in category/index.php
$id = $_GET['id'];
echo $id;
<a class='section_header' href="category/index.php?id=<?php echo $catid?>"><b><?php echo $row[0];?></b></a>
Try
<a class='section_header' href='category/index.php?id='" . $catid . "'><b>" . $row[0] ."</b></a>
You had a random ` in there. Once it prints out you can use inspect element to make sure the url looks correct and id is populated.
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I can not color the $_SESSION['username'] while echo in php.
echo "Welcome, " .$_SESSION['username']."! "; //this code is running
echo '<span style="color:#AFA;text-align:center;">Welcome,'$_SESSION['username'];'</span>'; // this part is not working...
Please help me if you can
You forgot the concatenation ..
echo '<span style="color:#AFA;text-align:center;">Welcome,' . $_SESSION['username'] . '</span>'; // this part is not working...
Notice the missing . before and after $_SESSION['username'];
echo'<span style="color:#AFA;text-align:center;">Welcome,'.$_SESSION['username'].'</span>';
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I am trying to display an image using a variable "$images" that contains the URL parsed form an API.
This is my code:
echo "<td>""<img src='",$image,"'>""</td>\n";
I assume there is a typo I cannot detect because I get a blank screen when running this.
echo "<td>"."<img src='".$image."'>"."</td>";
Or
echo "<td><img src='".$image."'></td>";
You were missing the dots/commas after/before the td tags
use . not ,
<img src='".$image."'>
PHP requires that you chain your strings together using a .
E.g.
echo 'Test' . ' ' . 'Hello'; // Test Hello
Or simply :
echo "<td><img src='$image'></td>";
Check documentation.
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Closed 8 years ago.
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It's a basic question, but I've googled all sorts of variations of it & nothing that quite answers me. I want to display one of two images. One if the criteria is met & one if it is not met. This code works. If criteria is met, it does display the image. But there is no alternative
<? if(stripslashes($getfeedbackQryRow['CompanyID'])=='344'){?><img src='img1.png' alt='Todays Bite'><? }?>
Then I put this together, but it's still wrong
<?PHP
if ($getfeedbackQryRow['CompanyID']) == '344') {
print ("<IMG SRC =/img1.png>");
}
else {
print ("<IMG SRC =img2.png>");
}
?>
This one looks like it should be right, but it throws an error... It's probably a syntax issue. Does anyone know what I'm doing wrong?
You have an extra ")"
if ($getfeedbackQryRow['CompanyID']) == '344') {
Remove the ")" here ['CompanyID'])
if ($getfeedbackQryRow['CompanyID']) == '344') {
// ^------- this is causing your error
You should do something simple like this:
$image_name = $getfeedbackQryRow['CompanyId'] == '344' ? 'img1.png' : 'img2.png';
echo '<img src="' . $image_name . '">';
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Closed 8 years ago.
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im getting this error but cannot figure out why
while($info = mysql_fetch_array( $data ))
{
echo itg_fetch_image('.$info['meta_value'].'); //the line giving error
Print "<tr>";
Print "<th>ID:</th> <td>".$info['meta_id'] . "</td> ";
Print "<th>VALUE:</th> <td>".$info['meta_value']. "</td> ";
Print "<th>DONE:</th> <td>YES</td> ";
}
if i comment out this line it shows fine in the value td, ive tried everthing. taking the dots out the taking the comments out then adding "". itg_fetch_image is http://www.intechgrity.com/automatically-copy-images-png-jpeg-gif-from-remote-server-http-to-your-local-server-using-php/#
the code is ment to be echo itg_fetch_image('url') and all meta_values return a url string
The quotes on the echo line are causing the problem.
It should look like this:
echo itg_fetch_image($info['meta_value']);
it seems that itg_fetch_image() is a function
try this:
echo itg_fetch_image($info['meta_value']);
you can just pass variable . do not need to use quotes.
You must write it as
echo itg_fetch_image($info['meta_value']);