Asked  7 Months ago    Answers:  5   Viewed   25 times

i want to echo out everything from a particular query. If echo $res I only get one of the strings. If I change the 2nd mysql_result argument I can get the 2nd, 2rd etc but what I want is all of them, echoed out one after the other. How can I turn a mysql result into something I can use?

I tried:

$results = mysql_query($query);
$res = mysql_result($results, 0);

while ($res->fetchInto($row)) {
    echo "<form id="$row[0]" name="$row[0]" method=post action=""><td style="border-bottom:1px solid black">$row[0]</td><td style="border-bottom:1px solid black"><input type=hidden name="remove" value="$row[0]"><input type=submit value=Remove></td><tr></form>n";


$result = mysqli_query($conn, $sql); // First parameter is just return of "mysqli_connect()" function
echo "<br>";
echo "<table border='1'>";
while ($row = mysqli_fetch_assoc($result)) { // Important line !!! Check summary get row on array ..
    echo "<tr>";
    foreach ($row as $field => $value) { // I you want you can right this line like this: foreach($row as $value) {
        echo "<td>" . $value . "</td>"; // I just did not use "htmlspecialchars()" function. 
    echo "</tr>";
echo "</table>";
Wednesday, March 31, 2021
answered 7 Months ago


echo $Result['StructuredXMLResume']['ContactInfo']['PersonName']['FullName'];

For the EmployerOrgName, as there's more than 1 of these entries, you'll need a foreach loop to get them all, like so:

foreach($Result['StructuredXMLResume']['EmployerOrg'] as $x){
    echo $x['EmployerOrgName'] . '<br />';

For the gender:

echo $Result['StructuredXMLResume']['ResumeAdditionalItems']['ResumeAdditionalItem'][0]['Gender'];

I can't say for sure if this is correct as the array is very diffifcult to read through but from what I can gather, this is correct.

The best way, for you, to find these things for yourself, is printing smaller sections of the array, for example:


With this, you'll see that there are other elements in there but you're not having the screen cluttered up by the rest of the array.

Saturday, May 29, 2021
answered 5 Months ago

you can't print the result from mysqli_query, it is mysqli_resource and for dumping the error you need to change mysql_error() to mysqli_error()

$username = "bob";
$db = mysqli_connect("localhost", "username", "password", "user_data");
$sql1 = "select id from user_information where username='$username'";
$result = mysqli_query($db, $sql1) or die(mysqli_error());
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { 
    echo $row['id'].'<br>'; 
Saturday, May 29, 2021
answered 5 Months ago

yo need create the user "pma" in mysql or change this lines(user and password for mysql):

/* User for advanced features */
$cfg['Servers'][$i]['controluser'] = 'pma'; 
$cfg['Servers'][$i]['controlpass'] = '';

Linux: /etc/phpmyadmin/

Tuesday, July 13, 2021
answered 4 Months ago
foreach ($array as $key => $val) {
   echo $val;
Thursday, August 12, 2021
answered 3 Months ago
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