Echo PHP value inside a jquery Function - php

Question

I am trying to assign a php echo value to an input generated from a jquery function. But so far no luck. It breaks the function and no results are displayed along with the input field. What is the proper way for this scenario to display php value inside the query function.
PHP
$tablename = "table";
$next_increment = 0;
//$qShowStatus = "SHOW TABLE STATUS LIKE '$tablename'";
$qShowStatusResult = $db_con->prepare("SHOW TABLE STATUS LIKE '$tablename'");
$qShowStatusResult->execute();
$results = $qShowStatusResult->fetchAll(\PDO::FETCH_ASSOC);
foreach($results as $value){
$next_increment = $value['Auto_increment'];
}
var nextAutoIncrement = '"'<?php echo $next_increment; ?>'"';
Jquery
newSection.children(':nth-child(1)').children(':first').attr('id', 'auto_id_' + newNum).attr('name', 'auto_id_' + newNum).val(nextAutoIncrement).hide();

Answer 1

Try this
<script language="javascript" type="text/javascript">
var nextAutoIncrement = '<?php echo $next_increment;?>';
</script>

Answer 2

Try like this:
<script language="javascript" type="text/javascript"
var nextAutoIncrement = <?php echo $next_increment; ?>;
</script>

Answer 3

<script>
//if it is anumber
var nextAutoIncrement = <?php echo $next_increment; ?>;
// if ity is a string
var nextAutoIncrement = '<?php echo $next_increment;?>';
</script>

Answer 4

In JS code there is require to define the <script> tag:
$tablename = "table";
$next_increment = 0;
$qShowStatusResult = $db_con->prepare("SHOW TABLE STATUS LIKE '$tablename'");
$qShowStatusResult->execute();
$results = $qShowStatusResult->fetchAll(\PDO::FETCH_ASSOC);
foreach($results as $value){
$next_increment = $value['Auto_increment'];
}
<script type="text/javascript" >
var nextAutoIncrement = '<?php echo $next_increment; ?>';
</script>