在“ for”循环的每次迭代中，如何增加mysql中的行？

Question

I want my little script to take a link that i have already inserted into the database, get therelevent text from that webpage and insert it into a different column on that same row. Then do it again on the next row. I am getting this "Object of class mysqli could not be converted to string" error but Im not sure how to represent the row number correctly. Heres what i have so far.

UPDATE Ok guys I appreciate your help. I think I unclear about this. The error message says it on the line that says $query1 = "$conn, SELECT adlink, key FROM usedcars WHERE key = $x"; Or at least, thats what its saying right now. I have tried all the solutions suggested, and so far I just get a bunch of other errors.

for ($y=1;$y=1201;$y++)
    {
    $x = 1;

    $query1 = "$conn, SELECT adlink, key  FROM usedcars WHERE key = $x";

    $query2 = "$conn, INSERT INTO usedcars (adtext), VALUES ($final_text) WHERE key = $x";

    $link_result = mysqli_query($query1);

    $text_holder = file_get_contents($link_result);
    $final_text = parse_array($text_holder, "postingBody", "<!-- .posting -->");

    mysqli_query($query2);
    echo "<font size='18' color='#FFFF00'>Placing text from $link_result into database</font><br>"; 
    $x++;
    }

I know its probably something simple...Im just not as smart as you guys.

Answer 1

first check whether all fields with key = $x exists

use update instead of insert

WHERE is used to filter existing results as said by @Eliel not to add a new one...

$query1 = "$conn, SELECT adlink, key  FROM usedcars WHERE key = $x";

$query2 = "$conn, UPDATE usedcars (adtext), VALUES ($final_text) WHERE key = $x";

$link_result = mysqli_query($query1);

if want to add a new one use:

$query1 = "$conn, SELECT adlink, key  FROM usedcars WHERE key = $x";

$query2 = "$conn, INSERT INTO usedcars (adtext,***,***) VALUES ($final_text,***,***)"; // provide all necessary fields

$link_result = mysqli_query($query1);

Answer 2

When you are using $query1 = "$conn, SELECT adlink, key FROM usedcars WHERE key = $x"; and pass it to mysqli_query() , the $conn is assumed as a string and it tries to use $conn as string.Thats why the error is coming.

try this -

$query1 = "SELECT adlink, key  FROM usedcars WHERE key = $x";

$query2 = "UPDATE usedcars (adtext), VALUES ($final_text) WHERE key = $x";

$link_result = mysqli_query($conn, $query1);

Answer 3

this is your query

for ($y=1;$y=1201;$y++)
{
$x = 1;

$query1 = "$conn, SELECT adlink, key  FROM usedcars WHERE key = $x";

$query2 = "$conn, INSERT INTO usedcars (adtext), VALUES ($final_text) WHERE key = $x";

$link_result = mysqli_query($query1);

$text_holder = file_get_contents($link_result);
$final_text = parse_array($text_holder, "postingBody", "<!-- .posting -->");

mysqli_query($query2);
echo "<font size='18' color='#FFFF00'>Placing text from $link_result into database</font><br>"; 
$x++;
}

My suggested query

 $x = 1;//put it before loop, every loop this $x will become one and the incrementation doesn't take effect
for ($y=1;$y=1201;$y++)
{


$query1 = "$conn, SELECT adlink, key  FROM usedcars WHERE key = $x";

$query2 = "$conn, INSERT INTO usedcars (adtext), VALUES ($final_text) ";//remove your where

$link_result = mysqli_query($query1);

$text_holder = file_get_contents($link_result);
$final_text = parse_array($text_holder, "postingBody", "<!-- .posting -->");

mysqli_query($query2);
echo "<font size='18' color='#FFFF00'>Placing text from $link_result into database</font><br>"; 
$x++;
}