Getting Illegal string offset Warning while fetching data

Question

$cq = "SELECT * FROM salescard WHERE custid = '$custid'";
$rq = mysqli_query($conn,$cq);
if ($c = mysqli_fetch_array($rq, MYSQLI_ASSOC)) {

$edate =$c['expdate'];
$ono = $c['orderno'];
$cno = $c['cardno'];
$c = $c['cvv'];
$email = $c['email'];
$no = $c['no'];

}

I am fetching data from the table and echoing the variables fetched above.

All the variables are echoed properly expect $email and $no.

Warning: Illegal string offset 'email' in 
C:\xampp\htdocs\erp\admin\viewsales\card.php on line 94

Warning: Illegal string offset 'no' in 
C:\xampp\htdocs\erp\admin\viewsales\card.php on line 95

All i know is this warning has something to do with variable being defined where an array is expected.

Only two variables are showing the error and rest are fetching properly How can i make this work properly?

EDIT

i surely have these two columns in my salescard table and i was able to store data into it from another form.

$email = $c['email'];
$no = $c['no'];

These are the lines 94 and 95

And i want to display the data here

<div class="form-group half left clear">
  <input type="text" value="<?php echo $email;?>" name="email" />
  <label class="control-label" for="input">Email Address</label><i 
class="bar"></i>
</div>

 <div class="form-group half right">
  <input type="text" value="<?php echo $no;?>" name="no" />
  <label class="control-label" for="input">Contact number</label><i 
class="bar"></i>       
 </div>

do you have email,no columns in your salescard table? Also where you are using these variables can you show that code (like 94,95 lines)? — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Jul 29, 2017 at 22:50

jh1711 · Accepted Answer · 2017-07-29 22:55:40Z

3

You overwrite $c in line 93, and make it a string.

$c = $c['cvv'];

Should be:

$cvv = $c['cvv'];

answered Jul 29, 2017 at 22:55

jh1711

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9 Comments

webnoob Over a year ago

This also stresses the importance of having meaningful variable names. $array or $data would suit so much more than $c.

Arun Over a year ago

damn!! these minor mistakes really costing me some time .. saving time with short variable names did backfired .. anyway thanks :)

Michael - sqlbot Over a year ago

@Arun it's pretty terrifying that you appear to be processing credit card data with code that looks lile like ... WHERE custid = '$custid'"; Variable interpolation into SQL queries is, for all practical purposes, never an acceptable practice. This code seems vulnerable to SQL injection. If you learned this from a book, you should set the book on fire.

jh1711 Over a year ago

@Michael-sqlbot. I'll admit that it is not clear who provides $custid. But without that information your comment seems harsh. If the source for $custid can't be trusted, and $custid is not sanitzed, you're absolutely correct.

Michael - sqlbot Over a year ago

@jh1711 agreed, it did sound harsh. It was a reaction to the notion of "saving time" (though I didn't explicitly say so) --building queries like this is "easy" but the danger of such practices can't really be overstated. Even concepts like "trusted" and "sanitized" are not really sufficient justifications for concatenation, as subsequent refactoring can undo those things and the undoing can go unnoticed. Good eye on the answer, btw. Seeing the magic string "cvv" was what initially set me off. CVV is never supposed to be stored in a database. You discard that after authorization.

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