hendrikbez Posted June 17, 2022 Share Posted June 17, 2022 I an new to this, still learning all the way. I have a table with more than 100 entries, each entry is different, but on one column I want to use 1 of three images to display in each entry. Want to do it in LEGIT, 3 images, red, yellow and green light to show This is my code, that I have, I do not know what more to give you PHP <?php if (isset($_GET['COIN_ID'])) { $COINGECKO_COIN_ID = $_GET['COIN_ID']; $sql = mysqli_query($conn,"SELECT * FROM coingecko_coin_custom_informaton WHERE COINGECKO_COIN_ID = '$COINGECKO_COIN_ID' "); $row = mysqli_fetch_array($sql); do{ $COINGECKO_COIN_ID = $row['COINGECKO_COIN_ID']; $COINGECKO_TOTAL_COIN = $row['COINGECKO_TOTAL_COIN']; $COINGECKO_LEGIT = $row['COINGECKO_LEGIT']; $COINGECKO_LINK = $row['COINGECKO_LINK']; }while($row = mysqli_fetch_array($sql)); } Here is my html <tr> <td><font color="#EFE1CE"><b><label for="firstNum1"><?php echo $coingecko_name; ?></label></b></FONT></td> <td style="color:#EFE1CE"><?php echo $COINGECKO_ID; ?></td> <td style="color:#EFE1CE"><?php echo $checkSqlRow["COINGECKO_LEGIT"]; ?></td> <td style="color:#EFE1CE"><?php echo $coingecko_rank; ?></td> Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/ Share on other sites More sharing options...
ginerjm Posted June 17, 2022 Share Posted June 17, 2022 (edited) Your html code: <?php echo " <tr> <td><b><label for='firstNum1'>$coingecko_name</label></b></td> <td style='color:#EFE1CE'>$COINGECKO_ID</td> <td style='color:#EFE1CE'>{$checkSqlRow["COINGECKO_LEGIT"]}</td> <td style='color:#EFE1CE'>$coingecko_rank</td> "; A few var names that are not defined (to us). And a bit cleaner just by using php to do your output. Now - what is the problem? You did not tell us. And where is your attempt to make some kind of selection of the image you want to be chosen? BTW. Note that I removed your use of the deprecated 'font' tag. Your use of CSS later on is much preferred nowadays. Edited June 17, 2022 by ginerjm 1 Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597397 Share on other sites More sharing options...
ginerjm Posted June 17, 2022 Share Posted June 17, 2022 I appreciate the 'like', but what is it that I/we can do for you? Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597400 Share on other sites More sharing options...
hendrikbez Posted June 17, 2022 Author Share Posted June 17, 2022 1 hour ago, ginerjm said: Your html code: <?php echo " <tr> <td><b><label for='firstNum1'>$coingecko_name</label></b></td> <td style='color:#EFE1CE'>$COINGECKO_ID</td> <td style='color:#EFE1CE'>{$checkSqlRow["COINGECKO_LEGIT"]}</td> <td style='color:#EFE1CE'>$coingecko_rank</td> "; A few var names that are not defined (to us). And a bit cleaner just by using php to do your output. Now - what is the problem? You did not tell us. And where is your attempt to make some kind of selection of the image you want to be chosen? BTW. Note that I removed your use of the deprecated 'font' tag. Your use of CSS later on is much preferred nowadays. Thank you I did not give my attempt, as I do not know on how to do the image (Have nor any idea on how to do it, as I am new to php, I did add a lot of new code to what I have received from a person . What i trying to get is, I have a html table, with more than 100 tokens in the table. 1 of my columns id for (legit), at the moment I type in "Yes" / "No" or "not at the moment"( easy to do for each one, when I add it in my admin file), bur I wnt to use 3 images on it (red/ green and yellow(this is where I do not know how to use it, how do add it in my admin file. So when I update any one of the tokens, I can use any on of the three images. Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597401 Share on other sites More sharing options...
ginerjm Posted June 17, 2022 Share Posted June 17, 2022 NO idea what you are asking. So far It's all about a table, not php. Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597402 Share on other sites More sharing options...
mac_gyver Posted June 17, 2022 Share Posted June 17, 2022 we don't know what your data is and don't know what result you want to produce from that data, so, your description doesn't make any sense. please post some sample data (not a picture of your data) and show or describe what result you want to produce from that data. as to the posted php code - validate all input data before using it. what should your code do if $_GET['COIN_ID'] isn't set, is empty, or doesn't contain an integer value greater than zero? since this is a required input for the code to work, you need to setup a user message stating that a required coin id was not supplied, and then not run the rest of the code that's dependent on that input. don't copy variables to other variables for nothing. this just a waste of your time typing. just use the original variables. don't put external, unknown, dynamic values directly into an sql query statement. use a prepared query instead. if this query will match at most one row of data, just directly fetch it, no loop is needed. if this query can match a set of zero or more rows of data, you would need to fetch the data into an array, then test/loop over the array to produce the output. the current code overwrites each row of fetched data with the next row, leaving you with only the last row of data. generally, do/while loops are not used since you must test if there's data to fetch at all, requiring more code. the current code will produce php errors if the query didn't match a row, and leave you needing to still add missing logic at the point of producing the output to display a message for the case where there is no matching data. Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597403 Share on other sites More sharing options...
hendrikbez Posted June 17, 2022 Author Share Posted June 17, 2022 6 minutes ago, ginerjm said: NO idea what you are asking. So far It's all about a table, not php. OK, will give the code, did not do changes yet as you said I must change. Let met try again to explain. See picture. I am now confused do I use php for it in the table (Legit) one. <table class="table-sortable mt32 customers-list" id="datatable"> <thead> <tr> <th height="50">Naam</th> <th height="50">Id</th> <th height="50">Bestaan Munt Nog</th> <th height="50">Rang</th> <th height="50">Skakel</th> <th height="50">Munt Skakel</th> <th height="50">Blokketting</th> <th height="50">Waar</th> <th height="50">Beursie Nommer</th> <th height="50">Beursie</th> <th height="50">Geverifieer</th> <th height="50">Prys</th> <th height="50">Dag</th> <th height="50">Week</th> <th height="50">Maand</th> <th height="50">Jaar</th> <th height="50">Munte</th> <th height="50">Usd</th> <th height="50">Zar</th> </tr> </thead> <tbody> <?php $total_multiply1 = 0; $total_multiply11 = 0; $total_multiply111 = 0; //Fetching CoinMarketCap table data if($fetchCoinMarketCapRowCount > 0){ foreach($decoded_json_coinMarketCap['data'] as $coinMarketCap) { // echo "<script>alert('HELLO');</script>"; $COINMARKETCAP_ID = $coinMarketCap['id']; $COINMARKETCAP_name = $coinMarketCap['name']; $COINMARKETCAP_symbol = $coinMarketCap['symbol']; $COINMARKETCAP_slug = $coinMarketCap['slug']; $COINMARKETCAP_name = htmlentities($COINMARKETCAP_name, ENT_QUOTES, "UTF-8"); $COINMARKETCAP_rank = $coinMarketCap['cmc_rank']; $COINMARKETCAP_price = $coinMarketCap['quote']['USD']['price']; $COINMARKETCAP_24h_change = $coinMarketCap['quote']['USD']['percent_change_24h']; $COINMARKETCAP_7d_change = $coinMarketCap['quote']['USD']['percent_change_7d']; $COINMARKETCAP_30d_change = $coinMarketCap['quote']['USD']['percent_change_30d']; $coinMarketCap24hColor = colorSet($COINMARKETCAP_24h_change); $coinMarketCap7dColor = colorSet($COINMARKETCAP_7d_change); $coinMarketCap30dColor = colorSet($COINMARKETCAP_30d_change); $checkSql = mysqli_query($conn,"SELECT * FROM coinmarketcap_coin_custom_info WHERE COIN_SYMBOL = '$COINMARKETCAP_symbol' ") or die(mysqli_error($conn)); $checkSqlRow = mysqli_fetch_array($checkSql); $checkSqlCount = mysqli_num_rows($checkSql); $COINMARKETCAP_price = sprintf('%.16f', floatval($COINMARKETCAP_price)); do{ $multiply111 = $COINMARKETCAP_price * $checkSqlRow["TOTAL_COIN"]; if($multiply111 != 0){ $multiply111 = sprintf('%.16f', floatval($multiply111)); } $total_multiply111 += $multiply111; ?> <tr> <td><font color="#EFE1CE"><b><label for="firstNum1"><?php echo $COINMARKETCAP_name; ?></label></b></FONT></td> <td style="color:#EFE1CE"><?php echo $COINMARKETCAP_symbol; ?></td> <td style="color:#EFE1CE"><?php echo $checkSqlRow["LEGIT"]; ?></td> <td style="color:#EFE1CE"><?php echo $COINMARKETCAP_rank; ?></td> <td> <a href="<?php echo $checkSqlRow["LINK"]; ?>" target="_blank"> <img src="<?php echo $decoded_json_coinMarketCap_logo["data"]["$COINMARKETCAP_symbol"]["logo"]; ?>" width="32px"></img> </a> </td> <td> <a href="<?php echo $checkSqlRow["SITE"]; ?>" target="_blank"> <img src="link.png" width="32px"></img> </a> </td> <td style="color:#EFE1CE"><?php echo $checkSqlRow["BLOCKCHAIN"]; ?></td> <td style="color:#EFE1CE"><?php echo $checkSqlRow["WHERE_NOW"]; ?></td> <td style="color:#EFE1CE"><?php echo $checkSqlRow["WALLET_NO"]; ?></td> <td style="color:#EFE1CE"><?php echo $checkSqlRow["WALLET"]; ?></td> <td style="color:#EFE1CE"><?php echo $checkSqlRow["VERIFIED"]; ?></td> <td id="firstNum2" style="color:#EFE1CE"><?php echo $COINMARKETCAP_price; ?></td> <td id="num1For24H" style="color:<?php echo $coinMarketCap24hColor; ?>"><?php echo round($COINMARKETCAP_24h_change,2); ?></td> <td id="num1For7D" style="color:<?php echo $coinMarketCap7dColor; ?>"><?php echo round($COINMARKETCAP_7d_change,2); ?></td> <td id="num1For30D" style="color:<?php echo $coinMarketCap30dColor; ?>"><?php echo round($COINMARKETCAP_30d_change,2); ?></td> <td id="num1For365D" style="color:white">NULL</td> <td id="secondNum2" style="color:#EFE1CE; font-weight:bold"><?php echo $checkSqlRow["TOTAL_COIN"]; ?></td> <td><input class="total-usd glow" type="text" id="<?php echo $COINMARKETCAP_symbol; ?>1" name="result5" value="<?php echo $multiply111; ?>" style="color:lime; background-color:black; font-weight:bold"/></td> <script type="text/javascript"> $(document).ready(function() { var COINMARKETCAP_symbol = '<?=$COINMARKETCAP_symbol?>'; var multiply111 = '<?=$multiply111?>'; var valueInZAR = $("#gcw_valFpWbgYyUL1").val(); var multiply222_total = multiply111 * valueInZAR; // alert(multiply22_total); if(multiply222_total!=0){ multiply222_total = multiply222_total.toFixed(10); } $("#"+COINMARKETCAP_symbol).val(multiply222_total); // alert(COINMARKETCAP_symbol); }); </script> <td><input class="total-usd glow" type="text" id="<?php echo $COINMARKETCAP_symbol; ?>" name="result6" style="color:lime; background-color:black; font-weight:bold"/></td> </tr> <?php Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597404 Share on other sites More sharing options...
ginerjm Posted June 17, 2022 Share Posted June 17, 2022 You do realize that php is case-sensitive and your usage of upper and lower variable names is gonna bite you right in the ass soon. Strongly suggest to stop. And now - good bye. You're on your own. You say you are still learning but you seem to have produced an awful lot of code so far. Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597405 Share on other sites More sharing options...
hendrikbez Posted June 17, 2022 Author Share Posted June 17, 2022 (edited) 4 minutes ago, ginerjm said: You do realize that php is case-sensitive and your usage of upper and lower variable names is gonna bite you right in the ass soon. Strongly suggest to stop. And now - good bye. You're on your own. You say you are still learning but you seem to have produced an awful lot of code so far. Yes thank you, did get this code from a professional php person ( he said he has written a lot for customers, and I have paid him for this work). Hope you feel better later on Edited June 17, 2022 by hendrikbez Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597406 Share on other sites More sharing options...
ginerjm Posted June 17, 2022 Share Posted June 17, 2022 So if you got this code from someone and if you don't know what it is, why aren't you asking him? Because he's not around anymore? Good luck then. IMHO he is not a very good programmer or he wouldn't have used mixed cases like that. Horrible practice. And certainly not such long variable names that make reading thru any code a tedious effort. And in your example, I don't see ANY images. And how do they fit into this grid/table of yours? Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597407 Share on other sites More sharing options...
mac_gyver Posted June 17, 2022 Share Posted June 17, 2022 this code is very bad, and doesn't even contain any comments, useful or otherwise. however, now that you have posted some information, your description makes sense. the wording for that column is in $checkSqlRow["LEGIT"]. is the stored wording consistent, i.e. are there only three values and are they always spelled and capitalized the same? next, for each possible wording, you need to map the value to the image path/name. the simplest way of doing this is to store this mapping in an array variable, with the wording as the array index, and the image path/name as the array value. at the current point of echoing $checkSqlRow["LEGIT"], you would instead test if the value/index is in the array. if it is, get the image path/name and output it in an <img ...> tag. if it is not, you would either output the wording, some default text, such as n/a, or output a default image. Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597408 Share on other sites More sharing options...
cyberRobot Posted June 17, 2022 Share Posted June 17, 2022 Looking at the screenshot and your code, it seems like the following variable is the one you will use to determine which of the 3 images to display. $checkSqlRow["LEGIT"] Your screenshot suggests this variable will contain one of the following values: weet nie nee ja One option would be to test for those values in an if / elseif statements to determine which image to display. For example, if the green image is to be displayed for "ja" and red for "nee", you could do something like this: <?php $checkSqlRow["LEGIT"] = strtolower($checkSqlRow["LEGIT"]); //make sure the value is lower case before testing if($checkSqlRow["LEGIT"] == 'ja') { echo '<img src="green.gif">'; } elseif($checkSqlRow["LEGIT"] == 'nee') { echo '<img src="red.gif">'; } elseif($checkSqlRow["LEGIT"] == 'weet nie') { echo '<img src="yellow.gif">'; } ?> Note that the image names (and the path) will need to be adjusted to whatever you are using. 1 Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597409 Share on other sites More sharing options...
hendrikbez Posted June 17, 2022 Author Share Posted June 17, 2022 2 hours ago, cyberRobot said: Looking at the screenshot and your code, it seems like the following variable is the one you will use to determine which of the 3 images to display. $checkSqlRow["LEGIT"] Your screenshot suggests this variable will contain one of the following values: weet nie nee ja One option would be to test for those values in an if / elseif statements to determine which image to display. For example, if the green image is to be displayed for "ja" and red for "nee", you could do something like this: <?php $checkSqlRow["LEGIT"] = strtolower($checkSqlRow["LEGIT"]); //make sure the value is lower case before testing if($checkSqlRow["LEGIT"] == 'ja') { echo '<img src="green.gif">'; } elseif($checkSqlRow["LEGIT"] == 'nee') { echo '<img src="red.gif">'; } elseif($checkSqlRow["LEGIT"] == 'weet nie') { echo '<img src="yellow.gif">'; } ?> Note that the image names (and the path) will need to be adjusted to whatever you are using. Thank you, for information, it is working Quote Link to comment https://forums.phpfreaks.com/topic/314937-how-to-show-1-of-three-pictures-as-image/#findComment-1597410 Share on other sites More sharing options...
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