PHP MYSQL array foreach and insert

Question

tbl1

id | detail
1  | this, that
2  | these, those

tbl2

id | visit  | cost
1  | this   | 20
2  | that   | 30
3  | these  | 40
4  | those  | 50

PHP:

<?php 

$id=$_GET['id'];    
$detail=$_GET['detail'];
$pieces = explode(",", $rows['detail']);

    // $pieces[0]; piece1
    // $pieces[1]; piece2        
    // $pieces[2]; piece2        
    // $pieces[3]; piece3    
    // $pieces[4]; piece4

    foreach($pieces[1] as $test){            

$query="SELECT * FROM tbl1, tbl2 WHERE
  $test = tbl2.cost AND
  tbl1.id='$id'";
    }   

$result=mysql_query($query) or die("error: " . mysql_error());

while($rows=mysql_fetch_array($result)){    
?>
<table>
    <tr>
    <td><input value="<? echo $rows['code'];?>" readonly="readonly"/></td>
    <td><input value="<? echo $rows['cost']; ?>" readonly="readonly"/></td>
    </tr>
  </table>

<?php
}
?>

for id=1, array 0=>this, 1=>that So, foreach($pieces[1] as $test), it means "that"

The above works, but, when I change $pieces[1] to $pieces[0], I still get the result which corresponds with $pieces[1]. And, when I change it to $pieces[4] which doesn't exist, I again get the result which corresponds with $pieces[1].

Where am I doing wrong? How can I make this array work?

According to tbl1.id=1 piece[0]=>this, piece[1]=>that and I want to make it work for my tables. Please help.

You have $pieces = explode(",", $rows['detail']); before you've done the database query that assigns to $rows. Should that be $pieces = explode(",", $detail);? — Barmar
– Barmar, Commented Aug 20, 2012 at 4:20

user557846 · Accepted Answer · 2012-08-20 03:38:33Z

1

$pieces[1], the variable you are using in your foreach loop is a string, not an array. You should be looping through the $pieces array so:

foreach($pieces as $test){

... NOT

foreach($pieces[1] as $test){

answered Aug 20, 2012 at 3:38

user557846

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7 Comments

user1605661 Over a year ago

error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= tbl2.cost AND tbl1.id='some numbers'' at line 3

user557846 Over a year ago

echo $query; what does that return?

user1605661 Over a year ago

return is : SELECT * FROM tbl1, tbl2 WHERE = tbl2.cost AND tbl1.id='some numbers' $pieces[1] is missing

user557846 Over a year ago

where is $rows['detail'] populated?

Tarun Over a year ago

try to echo only $pieces[1] only, also why are you comparing tbl2.cost(integer) with $pieces[1](string).

|

Tarun · Accepted Answer · 2012-08-20 03:41:12Z

1

$pieces[1] is not an array but an element of an array($pieces), so better use without loop statement,

$query="SELECT * FROM tbl1, tbl2 WHERE
  $pieces[1] = tbl2.cost AND
  tbl1.id='$id'";

Thanks

answered Aug 20, 2012 at 3:41

Tarun

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1 Comment

user1605661 Over a year ago

error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= tbl2.cost AND tbl1.id='some numbers'' at line 3

Steven Liao · Accepted Answer · 2012-08-20 03:45:50Z

0

When you run

$result=mysql_query($query) or die("error: " . mysql_error());

It will return a result of the from query. But what is your query? It is the last string you set $query to, which is the last time iteration of your for loop. That seems to be inefficient and not your intended behavior.

Assuming that you want all the results (and don't care about query optimization), you could do something like this:

foreach($pieces as $test){            
    $query="SELECT * FROM tbl1, tbl2 WHERE
    $test = tbl2.cost AND tbl1.id='$id'";

    $result[]=mysql_query($query) or die("error: " . mysql_error());
}

foreach ($results as $result){
    $rows=array_push(mysql_fetch_array($result));
}

Now, if you only want details from a specific piece, you don't even need a for loop.

// replace 3 with the desired element
$query="SELECT * FROM tbl1, tbl2 WHERE
$pieces[3] = tbl2.cost AND tbl1.id='$id'";

$result=mysql_query($query) or die("error: " . mysql_error());

mysql_fetch_array($result);

edited Aug 20, 2012 at 3:45

answered Aug 20, 2012 at 3:40

Steven Liao

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2 Comments

Steven Liao Over a year ago

Can you do a print_r on $pieces[3]?

user1605661 Over a year ago

empty result. I dont understand why?

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