PHP Fetch Data From Database - php

I'm trying to fetch data from hashtag where is #a2 or #ar hashtag but not work with my code
$sql_query = mysqli_query($conn, "SELECT * FROM posts WHERE hashtag=#a2");


try this,
$data = "#a2";
$sql_query = mysqli_query($conn, "SELECT * FROM posts WHERE hashtag LIKE '%$data%'");


$sql_query = mysqli_query($conn, "SELECT * FROM posts WHERE hashtag LIKE '%#a2%'");


You can also do it using your method but you just have to enclose #a2 in quotes like
$sql_query = mysqli_query($conn, "SELECT * FROM posts WHERE hashtag='#a2'");
Hope it helps

Related

How to search matching charecter in mysql

$result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE Sample_name LIKE '$userinput' OR Client_name LIKE '$userinput1%'") or die(mysqli_error($conn));
I am trying to execute above query, but its outputting full details from mysql database.
how to fix this issue?
thanks in advance!!

Missing % -
Sample_name LIKE '%$userinput%' OR Client_name LIKE '%$userinput1%'
Check the docs

You have to use '%' operator. A Like always expect a %. Try this -
$result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE Sample_name LIKE '%$userinput'") or die(mysqli_error($conn));

you missed % sign in $userinput
$result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE Sample_name LIKE '%$userinput%'") or die(mysqli_error($conn));

How to run an sql query inside a while loop of another query

The following works with no problem at all, when the photoId is directly on the statement and not a variable.
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "103"') or die(mysqli_error($con));
but the following just won't work with no error, what might be causing this not to select.
$imageid = '103';
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "$imageid"') or die(mysqli_error($con));
$img_row = mysqli_fetch_array($img_query);
echo $img_row['img'];
This is inside a while loop.
while($row = mysqli_fetch_array($somequery)){
$imageid = $row['photoid'];
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "$imageid"') or die(mysqli_error($con));
$img_row = mysqli_fetch_array($img_query);
echo $img_row['img'];
}
Thanks.

in php a ' and a " are very different and the query syntax is double quote around the query and single quote around variables.. although I would recommend you look at using parameters on your query instead of just putting a variable directly into the query
Per my recommendation you should change your query to this:
$imageid = '103';
$query = $con->prepare("SELECT * FROM imgs WHERE photoid = ?");
$query->bind_param('sssd', $imageid);
$query->execute();
this is just the nuts and bolts of it... if you want more information about the connection.. error handling and everything else read the DOCS

there is a big difference between ' and " in php
Differences
change your query to be
$img_query = mysqli_query($con, "SELECT * FROM imgs WHERE photoid = '$imageid'") or die(mysqli_error($con));
and it should work.

Issue with mysql query when changing value to a variable

$results = mysql_query("select * from doctorlist where assignednumber = '1231231234' ");
I need to change the number 1231231234 to a variable. If I change it to the code below it does not work. I have displayed the variable on the page so I know it is set.
$results = mysql_query("select * from doctorlist where assignednumber = '$phoneNumber' ");
Could someone please help. I know it is a small issue, but have been unable to fix it.

Perhaps split it like this
$sql_query = "select * from doctorlist where assignednumber='$phoneNumber'";
$results = mysql_query($sql_query);
or
$sql_query = "select * from doctorlist where assignednumber='".$phoneNumber."' ";
$results = mysql_query($sql_query);

First check your variable type with var_dump($phoneNumber) than do the following:
$results = mysql_query("select * from doctorlist where assignednumber = '".$phoneNumber."' ");
to improve readability and last if you expect an Integer cast your variable like:
(int)$phoneNumber
or if string do
mysql_real_escape_string($phoneNumber)

Try using the variable inside the query like this:
'{$phoneNumber}'

Using an array in a Where clause

So what I need to do is find in my tables where id = id and name = name. I have tried this with no success.
$sql="SELECT * FROM $mysqltable WHERE id='$id' && name='$name'"
EDIT I found that I was missing a } bracket below this and the above code does work :/ Thank you everyone.

This is how i would have done it: Also try entering the table name instead of a variable.
$sql = "SELECT * From $mysqltable WHERE id='$id' AND name='$name'";
$result = mysql_query($sql) or die(mysql_error());

Empty query result in php

I'm having troubles with a PHP code. The problems come when I execute a SQL query with a PHP variable inside it. The result doesn't shows anything, and the field in the database is not empty. Tried out with a static id (not variable) and the query works fine.
Do you know where I'm doing it wrong?
The query code:
$consultaVideoSeleccionado1 = mysql_query("SELECT * FROM videos WHERE idvideo = `'.$videoSeleccionado.'`;", $conexion);

Try this:
$consultaVideoSeleccionado1 = mysql_query("SELECT * FROM videos WHERE idvideo = `'.$videoSeleccionado.'`;", $conexion) or die(mysql_error());
That will give you an error message.
The problem is that you use both ` and ' as escape characters as the same time.
$consultaVideoSeleccionado1 = mysql_query("SELECT * FROM videos WHERE idvideo = '$videoSeleccionado';", $conexion);
should work.
Often things will be more clear when you echo the query so you can see the final result.

you are using double quotes on your query so there is no need for the dot . operator
$consultaVideoSeleccionado1 = mysql_query("SELECT * FROM videos WHERE idvideo = '$videoSeleccionado'", $conexion);

You are connecting string wrong! You are using ' single quote, but you should use double/none.
Try this query:
mysql_query("SELECT * FROM `videos` WHERE `idvideo` = '$videoSeleccionado';", $conexion)
Or
mysql_query("SELECT * FROM `videos` WHERE `idvideo` = '{$videoSeleccionado}';", $conexion)
Or
$q = "SELECT * FROM `videos` WHERE `idvideo` = '%s';";
mysql_query(sprintf($q, $videoSeleccionado), $conexion)
Edit:
If it's still not working problem can be in query, try checking if it is using mysql_error()(1) or try dumping query(2).
Example(1):
mysql_query("SELECT * FROM `videos` WHERE `idvideo` = '".$videoSeleccionado."';", $conexion) or die(mysql_error());
Example(2):
$q = "SELECT * FROM videos WHERE idvideo = '$videoSeleccionado';";
var_dump($q);
mysql_query($q, $conexion)

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